 So as we've been discussing L'Hopital's rule there's been a couple of indeterminate forms that we've investigated. We've talked about zero over zero and infinity over infinity where L'Hopital's rule applies directly. We've also talked about zero times infinity which can be turned into one of these quotient indeterminate forms for where L'Hopital's rule applies. It turns out there are several other indeterminate forms. There are three indeterminate forms that we want to talk about in this video which are of exponential type. They're going to look like zero to the zero infinity to the zero and one to the infinity. That is we're going to be taking limits of the form take the limit as x approaches a of f of x to the g of x where if we plugged a into f and into g we'll get things like zero to the zero infinity to zero and one to the infinity and each and every one of these indeterminate forms are in fact indeterminate forms because there's a there's a conflict of rules here right when you look at something like zero to the zero well when you take things to the zero power like you know zero square that should be zero if you like zero cube that should be zero so powers of zero should be zero but on the other hand if I take like two to the zero that's a one you get three to the zero that should be a one things to the zero power should be one so there's this conflict this clash of titans a who's more powerful a base being zero or an exponent being zero it turns out it's indeterminate because anything can happen what about the next one infinity to the zero power well if you take things like infinity squared that should be infinity infinity cubed that should be infinity but like we also mentioned you know things to the zero power should equal one so who's more dominant the base of infinity or the exponent of zero again we get this clash of titans these two powerful beast fighting for dominance and as such turns out it could be anything now this last one this last one is a lot harder for students to believe for some reason but it's the same conflict as before sure if you take things if you take powers of one like one squared one cubed one to the fourth these are always equal to one but what happens if you take things to the infinite power two to the infinity three to the infinity four to the infinity right these are typically infinite right in which case who's been more powerful the base of one or the exponent of infinity okay there's that conflict there and so who is who's the winner it could be anything so we need to investigate these things to be more careful now l'Hopital's rule only applies to forms zero over zero infinity over infinity like this zero times infinity thing we had to transform it into one of these quotient forms in order to apply l'Hopital's rule we have to do the same thing with these exponentials it's a little bit more tricky but this is the general strategy if you have an exponential expression what you're going to do is you're going to take the natural exponential in the log of of the expression here the reason we do that is because the exponential natural log or inverse operations they cancel each other out and so you get back f to the g now the advantage here is you take e to the natural log of f of f of x to the g of x power is because looking at just this piece right here the log of f to the g by logarithm properties the exponent comes out in front and this then becomes e e to the g times natural log of f power now if you take for example if f and g both go to zero if f and g both go to zero here then this thing looks like zero to the zero this thing looks like e to the natural log of zero to the zero but this thing right here is going to look like e times zero times the natural log of zero it's the natural log of zero is essentially negative infinity so you get e to the zero times infinity and this is important because this is now an indeterminate form that we know how to compute and therefore you can turn that to a product and that's how we're going to proceed to compute these exponential indeterminate forms they're a little bit more challenging because there's a lot of work going on here so consider this one at present the limit as x approaches zero positive for x to the x well if we just plug in zero we're going to get zero to the zero this is an indeterminate form so we have to investigate a little bit more so what we're going to do is we're going to take e to the natural log of x to the x this then becomes e to the x times the natural log of x and we want to take the limit of that so this limit is equal to the limit as x approaches zero from the right of e to the x times the natural log of x this is also another important step in this process here is that the exponential function is continuous so we can actually bring the natural exponential out of this function and we get something like e to the limit of x times the natural log of x as x goes to zero um sometimes because of the superscript notation it's a little bit more desirable to use instead of e you could say things like the function exp of x is equal to e to the x some people do that uh just so you can write it all in line so this is exp of the limit as x x times the natural log of x here as x goes to zero from the right because this has the form zero times infinity we need to switch this into a quotient and so we get exp of the limit of the natural log of x over one over one over x as x approaches zero and so now you see that this is going to have the indeterminate form infinity over infinity taking now we can apply l'opital's rule in which case we're going to get that exp of the limit here as x approaches zero from the right you're going to get one over x over negative one over x squared when you take the derivatives this fraction here is going to simplify just to be negative x so you're going to get exp of the limit as x approaches zero from the right of just a negative x so this is going to turn out to be exp of zero which exp reverses e to the zero power so we end up with just a one right here so the limit as x approaches zero from the right of x the x that is equal to one and so this then demonstrates exactly how we can compute these at these exponential internment forms we had three of them but the calculation of any one of them is really identical to the other ones it doesn't make much of a difference which form you have one to the infinity zero to the zero or infinity to zero the strategy is going to be all the same let's take a look at another such example let's take the limit as x approaches zero from the right of one plus sine of four x raised to the cotangent of x power notice that if we just plug in x equals zero we're going to get one plus sine of zero raised to the cotangent of zero power now sine of zero is zero so that base is just going to be one cotangent of zero is going to turn out to be infinity mostly basically we see that as you approach cotangent from excuse me approach zero from the right you find a vertical asymptote for cotangent so this is going to be one to the infinity is it one to the infinity or one to the negative infinity well we can investigate whether that is but it doesn't really matter the fact that cotangent has a vertical asymptote there is going to make the exponent go to the infinite and thus that activates this indeterminate form so what we need to then do is we need to compute the limit as x approaches zero from the right of e to the cotangent of x times the natural log of one plus sine of four x power that's the limit that we're going to have to compute next to go forward here but again as the e is continuous we can just pull out that exponential function so we get the limit as x approaches zero from the right of cotangent x times the natural log of one plus sine of four x and we can go on from here but like we saw in the previous example that the natural exponential e basically doesn't do anything until the very end of the problem so some people just kind of discard it and just focus on the limit as x approaches zero of cotangent x times the natural log of one plus sine of four x the idea is i'm just going to compute this and i'll just remember to do the e at the end but if you do that you have to actually remember to do the e it's a very common mistake to forget the e so this function now has the form as x approaches zero cotangent's going to go to plus or minus infinity we can figure out which direction it goes in but again the sine doesn't really matter here and in this one since sine is going to go to zero you're going to get the natural log of one natural log of one is zero this looks like infinity times zero we need to make this into a quotient the easiest way to make a quotient here i think is just to push cotangent into the denominator because cotangent has a very well known reciprocal function it's exactly just tangent and we can take the derivative tangent that seems a lot easier than pushing the natural log of one plus sine to the sine of four x into the denominator so this function now has the form of course that we're going to get what happens when we plug in x equals zero here we're going to get the natural log of one which is zero over tangent of zero which is zero itself it looks like zero over zero so low p tall's rule applies and so we're going to get the limit here as x approaches zero we're going to take the derivative of the natural log of one plus sine over four x that looks like well since it's a natural log you need a one plus sine of four x in the denominator then the inner derivative goes on top we're going to get four cosine of four x and the denominator when we take the derivative of c excuse me of tangent we're going to get a secant squared got a little ahead of myself there so that fraction's a little bit messed up let's simplify it we're going to get four cosine of four x over one plus sine of four x since we are dividing by a secant square that actually puts a secant squared on top but since secant itself is just a one over cosine i'm going to put that in the denominator so we ended with cosine squared of x like so taking the limit as x approaches zero now let's see what happens if we plug in x equals zero we would end up with a four times cosine of zero cosine of zero is one should be mentioned then you're going to get one plus sine of zero we've already observed that sine of zero zero and then you get cosine of zero which is itself one so we should get four times one over one times one this is going to be four now we might get so excited that we computed limit we have to remember the answer is not four the answer we act to find the answer we actually have to go back up to the white space above here right because we're we found out this limit is equal to e to the fourth that is the correct limit that's honestly in my opinion the most dangerous part of these exponential indeterminate forms we get so excited that we computed a limit we have to remember that the answer is going to be e to that number don't forget the natural exponential there uh e to the fourth is the correct limit calculation let's do one last example let's compute the limit as x approaches infinity of one plus three over x the x power notice what happens if we just plug in infinity we get one plus three over infinity raised to the infinite power this likewise looks like one to the infinity so what we're going to do is we need to compute the limit here as x approaches infinity of x times the natural log of one plus three over x so notice what I did there is I brought down the x point and slapped it in natural log to calculate this thing this function has the form infinity times the natural log of one which is zero times infinity that's always going to happen with these exponentials you go from an exponential to a product then we have to switch that into a quotient indeterminate form so I'm going to do that by pushing the x in the denominator take the limit as x approaches infinity of the natural log of one plus three over x and over one over x you rarely if ever put the natural log in the denominator here so now this function has the indeterminate form you'll notice the numerator will still look like the natural log of one which is a zero the denominator is going to look like one over infinity which is the same thing as zero so we have an indeterminate form that lopital's rule applies so by lopital's rule we're going to take the derivative top and bottom the derivative of the numerator well when it comes to derivatives of a natural log you always just put the original operand on the bottom then you have to take the derivative of the inner function that goes on top so you're going to get a negative three over x squared in the denominator we take the derivative one over x which is going to be negative one over x squared there is so much going on here again so many fractions since I'm dividing by a fraction I'm just going to write this as you know multiplying by the reciprocal I think that'll be an easy way to clean this thing up so we take the limit as x approaches infinity we get negative three over x squared over one plus three over x and then you're going to times this by the reciprocal which is going to be x squared over negative one we still have a bunch of nested fractions so what I'm going to do is I'm going to times x squared over x squared right here so that these guys cancel out and then you can distribute it looks like a three doesn't it let me make that look like a two and so then we're going to distribute this thing through what does this thing turn out to be take the limit again a lot a lot of mess here but we we multiply it through so in the new matter we're going to have a negative three x squared these x squareds cancels we still have the negative three in the x squared in the denominator when we distribute that negative x squared through we're going to get a negative x and a excuse me negative x squared we times it by the one and then we're going to get a negative three x like so what would happen here if we plug in x equals infinity we'd still get infinite right but in the case we're getting infinity over infinity that's exactly what I meant here we could do low details rule again but I'll notice that the top is a this is a rational function we have a negative three x squared on top we have a negative x square on the bottom since this is a balanced rational function as x goes to infinity this thing is going to turn out to be negative three over negative one we end up with just a three so there's other ways of calculating this thing but I'm going to I'm going to be content with that simplification going forward here and this then doesn't give us the final answer right our this this limit we did in yellow this is not equal to the original thing in fact what we have coming in over here that's a long equal sign right there what we get is that the limit is going to equal e to this number e cubed and so the final limits can be e cubed don't forget that e at the end because like we saw in the previous examples this stuff in yellow is just sort of scratch work to help us figure out this limit this limit is equal to three but this limit's not equal to the original limit will the original limit will equal to this limit raised by the base e