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Hand-in-Hand Teaching Guide to create Calculus II Met MA 124 A1 Form

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How to employ The Calculus II Met MA 124 A1 Form?

in this video we're going to go over.l'hopital's rule and the different.indeterminate forms you need to be.familiar with and we're going to go.through a ton of examples so you can.master this topic so let's begin let's.say if you want to find the limit as X.approaches a for a function let's say f.of X divided by G of X and let's say.that this function is in the form 0 over.0 or infinity over infinity these are.known as indeterminate forms because.what exactly is 0 over 0 0 over 5 for.example is 0 1 divided by 0 is undefined.so what exactly is 0 over 0 we don't.know infinity over infinity could be a.constant like 4 it could be 1 it could.be infinity it could be 0 we just don't.know what the value is so they're.indeterminate so let's say that the.limit as X approaches a for f of X is 0.and for G of X is 0 so we would have the.indeterminate form 0 over 0 if that's.the case or if we have infinity over.infinity then this limit is equal to the.limit as X approaches a F prime of X.divided by G prime of X and that's the.basic idea of l'hopital's rule if you.can't find the limit in this form take.the derivative of the numerator and the.denominator separately and you might be.able to find the derivative in this form.I mean the limit in that form now this.is going to make sense once we do a few.examples let's try an example what is.the limit as X approaches infinity for.the function e to the X divided by x.squared well let's see if we can use.direct substitution first so what is e.to the infinity e to a very large number.is going to be a large number it's.infinity.infinity squared is also infinity so.this is we what we have here is an.indeterminate form which means that we.can use l'hopital's rule to figure this.out because right now we don't know what.that is now if you plug in numbers you.can also find out the answer let's say.if we plug in a large number like 100 if.you type this in your calculator if you.can do it.e to the 100 divided by 100 squared is a.very large number it's two point six.eight eight times ten to the 39 direct.substitution is like your fallback.method anytime you are having difficulty.finding the answer this technique will.always work so if you plug this in this.large number represents infinity so.infinity is the answer but let's use.l'hopital's rule to confirm of that.answer so what we're going to do is.we're going to take the derivative of.the numerator and the derivative of the.denominator what is the derivative of e.to the X the derivative of e to the X is.simply e to the X and the derivative of.x squared is 2x now if we plug in.infinity at this point it's still going.to be infinity divided by infinity two.times infinity is infinity so we still.have the indeterminate form so we can.use l'hopital's rule a second time.the derivative of e to the X is e to the.X and the derivative of 2x is 2 so.notice that we no longer have the.indeterminate form if we replace X with.infinity it's going to be infinity.divided by 2 which is infinity and we.know that's going to be the final answer.so let's try another example.try this one what is the limit as X.approaches 0 sine X divided by X if we.plug in 0 we're going to have sine 0.divided by 0 sine 0 is 0 and 0 over 0 is.an indeterminate form we don't know what.that is equal to it could be 0 it could.be infinity they can be 5 it could be 1.we just don't know but we can use.l'hopital's rule but before we do that.let's plug in a number that's very close.to zero and let's see what the answer is.going to be so if you plug in a number.that's very very close to zero let's say.point zero zero one now this is of.course if you have access to calculator.when you're taking your test you can.always check your work.this is going to give you a small number.like point zero one seven now let's see.what happens if we plug in an even.smaller number let's say point is zero.zero zero zero one now there's two.things that can happen at this point.either the number could stay the same or.it can change if it stays the same then.that's going to be the limit if it gets.smaller then the limit is zero if it.gets larger than it's something else by.the way I do need to make a correction.and my calculator is in degree mode.someone in point zero one seven that's a.small number so don't make that mistake.make sure your calculator is in Radian.mode for this sort so sign point two.zero zero one divided by point zero zero.one well is actually 0.99 nine nine nine.so it's safe to say the limit is one if.you plug in sign point zero zero zero.zero one divided by that same number.again my calculators gives me one so we.know this is the answer but let's use a.l'hopital's rule to confirm that answer.so let's differentiate the numerator and.the denominator separately the.derivative of sine is cosine and the.derivative of X is 1 so we simply get.cosine X at this point we can plug in 0.cosine 0 is equal to 1 1 divided by 1 is.1.so as you can see l'hopital's rule it.works it will give you the right answer.as long as you have an indeterminate.form let's try another example the limit.as X approaches infinity for the.function Ln X divided by X so if we plug.in infinity at this point natural log of.infinity is infinity so we have the.indeterminate form infinity over.infinity which means that we can use.l'hopital's rule to get the answer now.if you think about these two functions.if you understand how this is going to.work like if you understand the concept.of l'hopital's rule you can pretty much.see what the answer is going to be for.instance before we do this one consider.this problem let's say if you have the.limit as X approaches infinity for the.function x squared plus 7x over X cubed.minus 8x the limit is basically the.horizontal asymptote notice that we have.a function that's a bottom-heavy the.degree of the denominator is higher than.that of the numerator X cube increases.at a faster rate than x squared.therefore the function is bottom-heavy.and whenever it's bottom a heavy the.answer is going to be 0 whenever it's.top-heavy it could be positive infinity.or negative infinity.now if the degree of the numerator is.the same as that of the denominator then.it's going to equal a constant like 5 a.negative 2.or something like that now which.function increases at a greater rate X.or LM X now if you were to plot the.function Ln X and X separately X.increases at a constant rate at a 45.degree angle the natural log of x.increases at a decreasing rate so X.increases faster than Ln X what that.means is that this function Ln X over X.is bottom-heavy since X increases at a.greater rate if the numerator increases.at a greater rate than the denominator.is going to be top-heavy in which case.the limit will be infinity or negative.infinity but because it's bottom-heavy.the final answer should be zero so let's.see if we can get that answer but first.let's plug in a number that's close to.infinity so let's plug in a thousand Ln.one thousand divided by a thousand this.is equal to point zero zero six nine.which is close to zero now let's try a.large number let's say Ln 1 million or 1.times 10 to the 6.this is equal to the point zero zero.zero zero one three eight so therefore.we could say that the limit is.approaching zero since it's bottom-heavy.x increases at a greater rate than LMX.now let's use l'hopital's rule to.confirm that answer so let's.differentiate the top and the bottom.part of the function the derivative of.the natural log function or Ln X is 1.over X and the derivative of X is 1 so.Ln I mean the limit as X approaches.infinity for 1 over X which is 1 over.infinity this is zero whenever you have.a large number on the bottom it's always.zero anytime you divide by a large.number you're going to get a smaller.number.now let's try another example the limit.as X approaches infinity for sine x over.X now this is not zero by the way this.is infinity and so when X approaches.infinity you need to consider the end.behavior of the function now sine X is.simply a periodic function it simply.repeats whereas the function for X it.increases so X increases at a greater.rate than sine therefore we can say the.function is bottom-heavy.this technique only applies if X.approaches a very large number either.infinity or negative infinity it doesn't.work if X approaches zero or some.constant so since we know that it's.bottom-heavy since X increases at a.greater rate than sine X we know the.final answer is going to be zero if you.plug in a large number into the equation.let's say a thousand make sure your.calculator is in Radian mode sine 1000.divided by a thousand is equal to point.zero zero zero eight three so we'll know.it works let's confirm the answer with.l'hopital's rule actually can we use.l'hopital's rule in this particular.problem.it turns out we can sign infinity.divided by infinity is not an.indeterminate form sign infinity can.vary between negative 1 to 1 divided by.infinity that's definitely going to be 0.anytime you have a constant divided by.infinity it's always equal to 0 so.l'hopital's rule doesn't work for this.equation we don't have an indeterminate.form nevertheless we can still figure.out what the answer is going to be you.can also use the squeezed term to prove.that this is equal to 0 let's try.another example what is the limit as X.approaches 2 for the function x squared.plus X minus 6 divided by x squared.minus 2x.so does l'hopital's rule work for this.particular example let's find out.let's plug in 2 so this is going to be 2.squared plus 2 minus 6 divided by 2.squared minus 2 times 2 2 squared is 4.and 2 times 2 is 4 4 plus 2 is 6 2 minus.6 is 0.so we have the indeterminate form 0 over.0 which means that l'hopital's rule can.work in this particular instance.sometimes it doesn't work so make sure.you have an indeterminate form before.using it now let's differentiate the top.and the bottom let's find the derivative.of x squared plus X plus 6 the.derivative of x squared is 2x the.derivative of X is 1 and the derivative.of the constant negative 6 is 0 now for.x squared minus 2x the derivative is.going to be 2x minus 2 so now let's plug.in 2 this is going to be 2 times 2 plus.1/2 times 2 minus 2.so 2 times 2 is 4 plus 1 is 5 2 times 2.is 4 minus 2 is 2 so this is the answer.it's 5 over 2 now another way in which.we can get that same answer is by.factoring notice that we have a.trinomial with a leading coefficient of.1 and the constant of negative 6 now.what are two numbers that multiply to.negative 6 but add to the middle term 1.this is going to be 3 and negative 2 so.to factor it it's going to be X plus 3.times X minus 2 to factor the bottom.part take out the GCF the greatest.common factor which is X and you'll be.left with X minus 2 notice that we can.get rid of the X minus 2 factor so what.we have left over is the limit as X.approaches 2 for the function X plus 3.divided by X and now we can use direct.substitution so 2 plus 3 divided by 2.which is 5 over 2 so we get the same.answer now let's try this example what.is the limit as X approaches infinity.for the function 5 minus X cubed divided.by x squared plus 4 let's make it plus X.cubed so the most significant term in.the numerator is X cubed and a.significant term in the denominator is x.squared X cubed increases at a faster.rate than x squared so therefore the.function is top-heavy which means that.the limit can be positive or negative.infinity it's probably going to be.positive infinity if we plug in infinity.right now this is going to be 5 plus.infinity to the third power divided by.infinity squared plus 4 which is.infinity over infinity so we can use.l'hopital's rule at this point so let's.find the derivative.of the top and the bottom the derivative.of 5 plus X cubed is 3x squared and the.derivative of x squared is 2x so right.now if we plug in infinity it's still.infinity over infinity so we can use.l'hopital's rule one more time the.derivative of 3x squared is 6x and then.the derivative of 2x is 2 so we'll now.have is the limit as X approaches.infinity for 3x which if we plug in.infinity 3 times infinity is infinity so.as you can see whenever you have a.top-heavy function or whenever the.numerator increases at a greater rate.than the denominator assuming if X.approaches a large number the answer.will always be infinity if it's.bottom-heavy where the denominator.increases at a greater rate Avandia.numerator when X is a large number then.it's going to be zero if it's.bottom-heavy try this one what is the.limit as X approaches infinity for the.function 2x squared minus 5x divided by.7 minus 4x squared so what do you think.the answer is going to be now the most.significant term in the numerator is x.squared or 2x squared now for x squared.let's ignore the negative sign but 4x.squared increases faster than 2x squared.in fact it increases twice as fast than.2x squared so the answer should be a.constant 2 or negative 2 probably.negative 2 because we have a positive.and a negative symbol either case 4x.squared increases twice as fast than 2x.squared so there's going to be a 2.somewhere in the answer now if we plug.in infinity this is going to be infinity./ negative infinity so this is.indeterminate so we can use l'hopital's.rule let's find the derivative of the.top and the bottom the derivative of 2x.squared is 4 X the derivative of 5x is 5.and for the constant 7 is 0 the.derivative of negative 4x squared is.negative 8x so this is what we have as.of now now if we plug in infinity into.this expression 4 times infinity is.infinity infinity minus 5 is still.infinity negative 8 times infinity is.simply a negative infinity so we have.another indeterminate form so we could.still use l'hopital's rule another time.the derivative of 4x minus 5 is simply 4.and the derivative of negative 8x is.negative 8 so this reduces to negative.1/2 which is the answer so as you can.see there's another way in which you can.get that same answer other than using.l'hopital's rule now it's important to.understand that the limit as X.approaches infinity for a constant.divided by X is equal to zero even if.you have a constant divided by x squared.it's still 0 when you take a finite.number like 5 12 20 divided by a very.very large number like infinity you're.gonna get a very very small number like.zero so knowing that let's multiply the.top and the bottom by 1 over x squared.since the degree of the numerator and.the denominator is 2 so 2x squared times.1 over x squared is simply 2 and 5 x.times 1 over x squared is 5 x over x.squared which is 5 divided by X and then.we're going to have 7 over x squared.minus 4 so the X Squared's will cancel.giving you negative 4 now we still have.the limit expression so the limit of a.constant divided by a variable when that.variable approaches infinity will be 0.so 5 over X and 7 over x squared will.turn into 0 so it's going to be 2 minus.0 over 0 minus 4 which is 2 divided by.negative 4 and you're going to get the.same answer negative 1/2.now keep in mind.a third way in which we can check the.answer is we can substitute infinity for.a very large finite number let's try a.thousand so what is a two times a.thousand squared minus five times a.thousand divided by 7 minus four times a.thousand squared a thousand squared is a.million times two that's two million.minus five thousand and then take that.result divided by seven minus four.million the answer is about negative.point four nine nine which is.approximately negative point five which.is very close to negative one half so we.could see negative one half is indeed.the right answer if you plug in ten.thousand you should get a number that's.even closer it's a negative point five.let's try this one what is the limit as.X approaches PI for the function sine X.divided by one minus cosine X so can we.use l'hopital's rule for this particular.example well let's find out let's.replace X with PI so what we have is.sine pi divided by one minus cosine PI.sine pi is equal to zero cosine pi is.negative one so this is zero over two.which is not an indeterminate form and.zero divided by anything is zero unless.it's zero over zero so the final answer.should be zero but let's confirm with.direct substitution so pi is about.3.14159 let's plug in 3.14 something.that's close to pi but that is not.exactly --part let's see if the answer.is going to approach.zero make sure you put this in.parentheses if you're going to type it.in your calculator sine of 3.14 divided.by one minus cosine 3.14 is about point.zero zero zero 796 which is close to.zero so we can say that answer is.acceptable now what about this one what.is the limit as X approaches infinity.for the function square root x over e to.the X what do you think the answer is.going to be well if we analyze the two.functions graphically the square root of.x looks like this it's an increase in.function but it increases at a.decreasing rate e to the X increases at.an increase in rate so e to the X.increases faster then square root X that.means that the function is bottom-heavy.and if X approaches a large number like.infinity for this particular situation.we know the final answer is going to be.zero so by analyzing the relative rates.of the top function and the bottom.function you can determine what the.answer is going to be if it's going to.be zero if it's going to be a constant.or plus or minus infinity now let's use.direct substitution to confirm that.answer so let's plug in a large number.like a thousand the square root of a.thousand divided by e raised to the 1000.okay my calculators given me an error e.to the learn thousand is just too big.let's try 100 if you try 100 this is.three point seven two times ten to the.minus 43 that's like point zero zero.zero zero zero zero forty something.times and then three seventy two so this.is approximately zero now let's see if.l'hopital's rule if it's going to work.let's substitute x with infinity so the.square root of infinity divided by e.raised to the infinity is infinity over.infinity which means that l'hopital's.rule should work so let's find the.derivative of the top and at the bottom.the square root of x function is x to.the one-half and the derivative of x to.the 1/2 is 1/2 x to the minus 1/2 and.the derivative of e to the x is going to.be the same thing which is e to the X so.right now we have a function that looks.like this 1/2 square root X e to the X.and of course listen off again the limit.expression which should be right in.front of it so now let's plug in.infinity into the equation so 2 1/2.square root infinity e to the infinity.which is 1 over infinity times e to the.Infinity a large number times another.large number is equal to a large number.so infinity times e to the Infinity is.going to be infinity and as we said.before whenever you have a constant or a.finite number divided by infinity it.will always equal 0 so that's the answer.for this particular problem.what is the limit as X approaches 1 for.the function X to the knife minus 1.divided by X raised to the fifth power.minus 1 so if we try to plug in 1 at.this point 1 to the 9th power is 1 and 1.minus 1 is 0 and the same is true for.the denominator so we have another.indeterminate form which means that we.can use l'hopital's rule so therefore.let's find the derivative of the top.part and the bottom part of the function.the derivative of x to the 9th power is.9 X to the 8th power and the derivative.of the constant 1 is 0 and 4x to the 5th.is simply 5 X to the 4th at this point.we can plug in 1 1 to the 8th power is.simply 1 so this is 9 times 1 on the.bottom 5 times 1 so the final answer is.9 over 5 which is 1.8 now let's check.our work by using direct substitution.let's plug in a number that's very close.to 1 so if we plug in one point zero.zero 1 raised to the ninth power minus.one divided by one point is zero zero 1.raised to the fifth power.you should get this number one point.eight zero three six which is.approximately one point or nine over.five so this is the answer try this.problem what is the limit as X.approaches zero from the right side of.the function X Ln X so what techniques.can we use to evaluate the limit if we.plug in zero a lanlan zero does not.exist an ln of a number close to zero.let's say like as X approaches zero from.the right that's going to be negative.infinity negative infinity times 0 is.also indeterminate which means we could.use l'hopital's rule but let's use.direct substitution first what's going.to happen if we plug in a number that's.just above zero or to the right of zero.if we plug in zero one I mean zero point.two zero one.what are we going to get if you type in.point zero one Ln point zero one this is.going to give you negative point zero.four six it's a small negative number.now if we plug in that number that's.even closer to zero like a point zero.zero one notice what happens.this is going to give us negative point.zero zero six line one so the answer is.still negative but it's approaching zero.it's getting smaller so to speak.therefore we know that the final answer.should be equal to zero but now how can.we use l'hopital's rule to get the same.answer whenever you have two functions F.times G multiplied to each other you.want to invite them as a fraction so you.want to convert the products the product.of F and G into a quotient and you could.do it by writing F / 1 / g f / 1 / g is.the same as f times g if you multiply.top and bottom by g these two will.cancel and you simply get F times G so.let's rewrite the expression X Ln X as.Ln X / 1 / X.now let's find the derivative of Ln X.and 1 over X the derivative of Ln X is 1.over X the derivative of 1 over X 1 over.X is the same as X 2 minus 1 and its.derivative is negative 1 X and minus 2.and when you rewrite it it's negative 1.over x squared so now let's simplify.this expression by multiplying the top.and the bottom part of the fraction by x.squared so these two will cancel and.we're going to get the limit as X.approaches 0 from the right side x.squared divided by X is simply X and x.squared divided by negative x squared is.simply negative 1 now we can plug in 0.at this point so the final answer is 0.now what about this one what is the.limit as X approaches negative infinity.x squared e to the X so if we use direct.substitution this is going to be.negative infinity squared e to the.negative infinity negative infinity.squared is going to be positive infinity.and this part we can move to the bottom.either the negative infinity which will.be e to the positive infinity e to the.positive infinity is infinity and so.infinity over infinity is indeterminate.which means we can use l'hopital's rule.now let's plug in a large number just to.see what the answer is going to be so.let's try negative 100 to represent.negative infinity so negative 100.squared e to the negative 100 if you.type this in.you will get three point seven two times.ten to the minus forty which is a very.very small number so basically the limit.equals zero now let's confirm the answer.with l'hopital's rule so somehow we need.to take this function and turn it into a.quotient fortunately a simple way in.which we could do that is e to the X is.1 divided by e to the negative X.whenever you change an exponent from.positive to negative you can move it.from the top to the bottom so we can.rewrite this as limit as X approaches.negative infinity x squared divided by e.to the negative x so now let's find the.derivative of the top and the bottom so.the derivative of x squared is 2x and.the derivative of e to the negative x is.e to the negative x times negative 1 so.now let's find the derivative one more.time.because if we plug in negative infinity.at this point it's going to be negative.infinity on the top e to the positive.Finity times negative 1 which is.negative infinity over negative infinity.and we still have an indeterminate form.so we can take the derivative second.time the derivative of 2x is simply 2.the derivative of negative e to the.negative x is the same thing times.negative 1 so if we substitute negative.infinity at this point these two.negative signs will cancel that's.positive 1 and then it's going to be e.to the negative negative infinity which.is 2 e to the infinity which is 2.divided by infinity whenever you have a.constant divided by infinity the final.answer is 0.how about this one what is the limit as.X approaches 0 from the right sign X Ln.X so let's see use direct substitution.as X approaches 0 from the right sign 0.is 0 and Ln X will become negative.infinity if you graph the natural log.function as you approach 0 from the.right side this function goes down.towards negative infinity so 0 times.infinity is indeterminate which means.we'll have to use l'hopital's rule so.let's rewrite it as limit as X.approaches 0 from the right Ln X divided.by 1 over sine X now what is the.derivative of Ln x + 1 over sine X for.Ln X it's simply 1 over X now for 1 over.sine X that's going to require some work.1 over sine X is sine X raised to the.minus 1 and if we use the power rule.it's going to be negative 1 sine X to.the minus 2 and according to the chain.rule we need to differentiate the.function inside the derivative of sine.is cosine so if we rewrite it it's going.to be negative cosine divided by sine.squared since we have a negative.exponent we can make it positive by.moving the sine squared to the bottom.now what should we do next what would.you do at this point let's multiply the.top and the bottom part by sine squared.or we really don't need to perhaps.you've heard of the expression keep.change flip so we can keep the top part.1 over X change division to.multiplication so times and then flip.the bottom part instead of cosine over.sine squared it's going to be negative.sine squared divided by cosine and of.course we still have the limit as X.approaches 0.so I'm going to rewrite the expression.the limit as X approaches zero from the.right and let's rearrange in sine.squared is sine times sine so I'm going.to take one of the sine functions and.move it here so it's going to be.negative sine x over X times sine x over.cosine X sine x over cosine X is tangent.X so we have is negative sine x over x.times tangent X perhaps you have seen.this formula earlier when you first.learn limits the limit as X approaches.zero for sine X divided by X is equal to.one so using that rule this portion will.become one so it's going to be negative.one times as X approaches zero tan 0 is.zero so it's negative one times zero.which is zero so that's the final answer.for this problem let's try this problem.what is the limit as X approaches.infinity for the function X minus Ln X.so if we plug in infinity infinity minus.Ln infinity is infinity minus infinity.which is indeterminate now if we plug in.then a large number to represent X let's.say a hundred 100 minus natural log of.100 is about.ninety-five point four which is a large.number if we plug in a thousand it's.going to be nine hundred ninety three.point one so we can see that as X.approaches infinity the limit approaches.infinity now let's prove it using.l'hopital's rule so if you have a.difference of two functions you need to.turn it into a fraction so we can put.this over one and let's let's multiply.top and bottom by the conjugate in this.case x plus lnx now if we foil x times X.is x squared and then x times Ln X.that's plus X Ln X and then negative Ln.X times X that's a negative X Ln X and.the last one Ln x times Ln X is Ln x.squared divided by X plus Ln X let's not.forget to write the limit expression so.the middle terms will cancel and now we.have the limit as X approaches positive.infinity x squared minus Ln x squared.divided by x plus lnx so if we plug in.infinity at this point it's still going.to be infinity minus infinity over.infinity plus infinity so it's still in.the indeterminate form so now let's take.the derivative of the top and the bottom.the derivative of x squared is 2x and.the derivative of ln x squared is 2.times Ln X raised to the first power.times the derivative of the inside.Ln part which is 1 over X the derivative.of X is 1 and the derivative of Ln X is.1 of X now let's multiply the top and.the bottom part by X 2x times X is 2x.squared and x times 1 over X they will.cancel so we're left with 2 Ln X on the.bottom x times 1 is X and 1 over x times.X is simply 1 now we could still use.l'hopital's rule again if we substitute.infinity for X is going to be infinity.minus infinity divided by infinity plus.1 so infinity minus infinity is.indeterminate so now let's take the.derivative of the top and the bottom.once more so we're going to have the.limit as X approaches infinity the.derivative of 2x squared is 4x the.derivative of Ln X is 1 over X and the.derivative of X is 1 now at this point.we can use direct substitution so this.is going to be 4 times infinity minus 2.times 1 over infinity 4 times infinity.is infinity and 1 divided by infinity is.0 so this is infinity minus 0 which is.infinity so this is the final answer now.what is the limit as X approaches zero.for the function cosecant x minus.cotangent X so feel free to try this one.now the first thing I would do is.rewrite it cosecant is simply 1 over.sine.and cotangent is cosine divided by sine.now whenever you have a difference of.two functions F minus G you want to.combine it into a single fraction.fortunately we already have the same.denominator so we can write this as the.limit as X approaches zero one minus.cosine X divided by sine X now let's see.if we can substitute zero at this point.if we do so it's 1 minus cosine zero.divided by sine zero cosine zero is.equal to 1 and sine zero is zero so zero.over zero we have an indeterminate form.which means that we can use l'hopital's.rule so let's go ahead and do that.so let's take the derivative of the top.and the bottom.the derivative of 1 is 0 and the.derivative of cosine is negative sine.and the derivative of positive sine is.cosine so sine divided by cosine is.tangent so the limit as X approaches 0.tangent X is equal to tangent 0 +.tangent 0 is 0 so 0 is the final answer.for this particular problem now let's.try another example what is the limit as.X approaches infinity for this function.square root x squared + 4 X - X so if we.plug in infinity the square root of.infinity squared plus 4 times infinity.minus infinity this is just going to be.infinity minus infinity and we don't.know what that is equal to so that's.another indeterminate form now let's.plug in a large number if we plug in 100.let's see what is going to happen.so if you plug in 100 you're going to.get 1.98 as an answer now let's increase.the value of x let's try a thousand.you.in this case is going to be one point.nine nine eight notice that as x.increases its approaching two so the.limit is equal to two now how can we.prove it what techniques can we use to.get the answer first.let's combine it into a single fraction.so right now the denominator is one so.we can just write it like this if we.want to now what I'm going to do is.multiply the top and bottom by the.conjugate of the numerator and that is.the square root of x squared plus four X.instead of minus X is going to be plus X.now whatever you do to the top.you must also do to the bottom so if we.foil the numerator the two middle terms.will cancel the square root of x squared.plus 4x times itself it's just going to.be x squared plus 4x and for the last.two negative x times X is negative x.squared and on the bottom it's going to.be square root x squared plus 4x plus X.the limit as X approaches infinity so.the terms x squared will cancel and.we're going to have the limit as X.approaches infinity for X on top./ square root x squared plus 4x plus X.now if we plug in infinity we're going.to have infinity over infinity which.means that we can use l'hopital's rule.but i've tried using l'hopital's rule.and for this particular problem it seems.like it's a never-ending process so I'm.going to do something else I'm going to.multiply the top and the bottom part of.the fraction by 1 over X notice what's.going to happen if I do that so I'm.going to have the limit as X approaches.infinity 4x times 1 over X is simply 4.divided by 1 over X square root x.squared plus 4x plus 1 now how can we.put the 1 over X inside the square root.it turns out that to do it it's going to.be square root 1 over x squared times x.squared plus 4x now here's why 1 over X.is the same as the square root of 1 over.x squared so we can replace 1 over X.with the square root of 1 over x squared.times x squared plus 4x so then we can.multiply 1 over x squared by x squared.plus 4x and that's going to be square.root 1 plus 4x divided by x squared.which is simply 4 over X.so we'll now have is the limit as X.approaches infinity 4 divided by square.root the x squared which will be.multiplied by 1 over x squared inside.the radical it's going to turn to 1 and.4x times 1 over x squared inside the.radical will be 4 over X now at this.point we can use direct substitution.de limit as X approaches infinity for.constant / variable this is going to be.4 over infinity which is equal to zero.so therefore this is going to become 4.divided by square root 1 plus 0 plus 1 1.plus 0 is 1 and the square root of 1 is.1 and 1 plus 1 is 2 and 4 divided by 2.is equal to 2 so the final answer is 2.what about something that looks like.this the limit as X approaches infinity.1 plus 2 over X raised to the X so if we.plug in infinity this is going to be 1.plus 2 divided by infinity is 0 raised.to the Infinity 1 raised to the Infinity.we don't know exactly what that's going.to be that's another indeterminate form.now let's find the answer by choosing a.very large value for X let's start with.100 1 plus 2 divided by 100 and raise to.the 100 and that's going to be 7 point 2.4 now let's try a thousand.one plus two divided by thousand raised.to the 1000 is about seven point three.seven now when you see equations that.look like this.typically it's associated with e the.answer is probably a squared e squared.is equal to a seven point three eight.nine so notice that it approaches a.squared but let's see if we can use.l'hopital's rule to get that answer.so for this problem we need to set the.entire thing equal to Y and our goal is.to solve for y whenever you have a.variable in the exponent position you.want to take the natural log of both.sides in order to bring it down so ln y.is equal to the limit as X approaches.infinity natural log 1 plus 2 over X.raised to the X so we can take this.variable and move it to the front that's.a property of logs so now we have the.limit as X approaches infinity X Ln 1.plus 2 over X so we have a product of.two functions f times G so we need to.turn into a quotient so instead of.having ln times x it's going to be Ln.divided by 1 over x so Ln Y is equal to.the limit as X approaches infinity.natural log 1 plus 2 divided by X.divided by 1 of X now let's find out if.we can use l'hopital's rule so if we.plug in infinity this is going to be Ln.1 plus 2 divided by infinity over 1.divided by Finity so 2 over infinity is.0 so we have Ln 1 1 divided by infinity.is 0 and Ln 1 is 0 0 over 0 is.indeterminate so we can use l'hopital's.rule so let's take the derivative of the.top and of the bottom the derivative of.a natural log function will say the.derivative of Ln U it's going to be u.prime divided by u so if u is equal to 1.plus 2 divided by X u Prime the.derivative of 1 is 0 the derivative of 2.X to the minus 1 is negative 2 X to the.minus 2 which is the same as negative 2.over x squared and the derivative of 1.over X which is X to the minus 1 that's.going to be negative.X.- - which is negative 1 over x squared.so what we have now is ln y is equal to.the limit as X approaches infinity.negative 2 divided by x squared so.that's the you prime part divided by U.which is 1 plus 2 over X so all of this.is the derivative of the Ln function and.then this is divided by the derivative.of 1 over X which is negative 1 over x.squared so keep in mind this is the U.prime part and this is the U part for.the natural log function so now we need.to simplify this particular expression.so what can we do here so first let's.multiply the bottom and the top by x.squared so these two will cancel and.that's simply going to be 1 x squared is.the same as x squared over 1 so negative.2 over x squared times x squared the X.Squared's will cancel and you're just.going to get negative 2 so what we have.is Ln Y is equal to the limit as X.approaches infinity negative 2 divided.by 1 plus 2 over X now let's not forget.this negative sign so these two negative.signs will cancel so we can make this.positive 2.now let's multiply top and bottom by X.so if we actually let's not do that.because we can evaluate the limit at.this point the limit as X approaches.infinity for 2 over X becomes 2 divided.by infinity so this is 2 over 1 plus 2.divided by infinity and 2 divided by.infinity is equal to 0 so this is 2 over.1 which is 2 so ln y is equal to 2 now.remember y is equal to our original.limit the original problem that we have.the limit as X approaches infinity for 1.plus 2 over X raised to the X so our.goal is to solve for y now the natural.log function has the base e so convert.it to its exponential form.AE raised to the second power is equal.to 1 so therefore the answer is e.squared which is what we predicted and.begin let's try another problem like.that let's evaluate the limit as X.approaches 0 1 minus 3x raised to the 1.over X so this is going to be something.associated with e again if we plug in 0.this is going to be a 1 minus 0 raised.to the 1 of something close to 0 which.is probably 120 with the infinity power.1 divided by a small number is going to.be a large number so we have another.indeterminate form which means that we.can use l'hopital's rule now let's set.the entire function equal to 1 now let's.see what the answer is going to be using.direct substitution.so let's plug in a number that's very.close to zero let's say a point is 0 0 1.so you should get point zero four nine.five six which is a small number now due.to the three is probably e to the third.but I think it's going to be e to the.negative three e to the negative three.is about point zero four nine seven.eight seven now let's see what's going.to happen if we plug in a number that's.even closer to zero so let's try point.zero zero zero one.you.this is going to be point zero four nine.seven six four notice that it's.approaching e to the negative three so.let's use l'hopital's rule to prove that.the limit is equal to e to the negative.three so let's begin by taking the.natural log of both sides.and let's move the exponent to the front.so it's one over x times ln 1 minus 3x.the good thing is we already have a.fraction which is going to make this.problem a lot easier so we can rewrite.it like this.this is ln 1 minus 3x divided by x so if.we plug in 0 at this point it's going to.be ln 1 minus 3 times 0 which is just 1.minus 0 divided by 0 ln 1 is 0 so we.have the indeterminate form which means.we can use l'hopital's rule so let's.take the derivative of the top and of.the bottom so remember the derivative of.a natural log function is equal to u.prime over u so u is the inside part of.the Ln function that's 1 minus 3x so u.prime is the derivative of 1 minus 3x.which is simply negative 3 so the.derivative of the natural log function.is negative 3 divided by 1 minus 3x.so this is going to be negative 3 over 1.minus 3x and the derivative of X is.simply 1 which we could ignore the 1 so.all we have is this so notice that we.can plug in 0 at this point so it's.going to be negative 3 over 1 minus 3.times 0 3 times 0 is simply 0 so what we.have is Ln Y is equal to negative 3.divided by 1 or simply negative 3 now.the base is e so to solve for y it's.going to be e raised to the negative 3.is equal to 1 and so since Y is equal to.the original problem the answer is e to.the negative 3.

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