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I have to solve this limit without using l'Hospital rule, the answer is $1/2$:

$$\lim_{x\to 0} \frac{\tan(x^2)}{(e^{2x} - 1)\sin x}$$

I have simplified it to this but don't know what to do next:

$$\lim_{x\to 0} \frac{x}{e^{2x} - 1}$$

tomnookey
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4 Answers4

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Without knowing what $e^x$ is as a Taylor series, we're going to struggle. $e^x$ can also be defined in other ways (e.g. as a limit), but it would be hard to evaluate this limit using those definitions. So let's take for granted:

$$e^{x}=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots+\frac{x^n}{n!}+\cdots $$

Now, let's substitute $x$ out for $2x$:

$$e^{2x}=1+2x+\frac{(2x)^2}{2}+\frac{(2x)^3}{6}+\cdots+\frac{(2x)^n}{n!}+\cdots $$

... and subtract $1$ so we get the denominator of our limit:

$$e^{2x}-1=2x+\frac{(2x)^2}{2}+\frac{(2x)^3}{6}+\cdots+\frac{(2x)^n}{n!}+\cdots $$

It would be great if there was a factor of $x$ I could take out of this, to cancel with the numerator...

$$e^{2x}-1=x\left (2+\frac{2^2x}{2}+\frac{2^3x^2}{6}+\cdots+\frac{2^nx^{n-1}}{n!}+\cdots \right)$$

So my limit is now:

$$\lim_{x\to 0} \frac{x}{e^{2x} - 1} =\lim_{x\to 0} \frac{x}{x\left (2+\frac{2^2x}{2}+\cdots+\frac{2^nx^{n-1}}{n!}+\cdots \right)}=\lim_{x\to 0} \frac{1}{2+\frac{2^2x}{2}+\cdots+\frac{2^nx^{n-1}}{n!}+\cdots }$$

As $x\to 0$, what happens to each of the terms in the denominator?

A.M.
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    I see it now. Thank you very much for the detailed answer, it helped me a ton. As x nears 0 each of the terms in the denominator nears 0 which leaves the answer, 1/2. Once again, thanks so much for you help! – tomnookey Feb 09 '22 at 17:14
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You're at stake. The limit $$ \lim_{x\to0}\frac{e^{2x}-1}{x} $$ is the definition of derivative of $f(x)=e^{2x}$ at $0$. You want the reciprocal of this limit, of course.

There is no way out: either you use a Taylor expansion, but this is nothing else than a generalization of l'Hôpital (or of Cauchy mean value theorem, if you prefer) or some “known” limit. In this case the “known” limit is $$ \lim_{x\to0}\frac{e^{x}-1}{x}=1 $$ and you can go to this form by doing $$ \lim_{x\to0}\frac{e^{2x}-1}{x}=\lim_{x\to0}2\frac{e^{2x}-1}{2x}=2 $$ So your complete derivation would be, using only “known” limits, $$ \lim_{x\to0}\frac{\tan^2x}{(e^{2x}-1)\sin x}=\lim_{x\to0}\frac{1}{2\cos^2x}\frac{\sin x}{x}\frac{2x}{e^{2x}-1}=\frac{1}{2} $$

egreg
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    Thank you very much for another answer! I'm pretty sure I was meant to use the Taylor series in this one but your way is very interesting as well. – tomnookey Feb 10 '22 at 09:45
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$$\lim_{x\to 0} \frac{x}{e^{2x} - 1}=\lim_{x\to 0} \frac12\frac{2x}{e^{2x} - 1}=\frac12$$

using the standard limit $\lim_{x\to0}\frac{e^x-1}{x}=1$

DatBoi
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    Could you cite or explain how $\lim_{x\to0}\frac{e^x-1}{x}=1$ is correct? Because it returns an indeterminate form as far as I can tell, not 1. – tomnookey Feb 09 '22 at 17:07
  • Welcome to the site @tomnookey! This is a standard limit which can be solved in a multitude of different ways, as shown here!! – DatBoi Feb 09 '22 at 17:09
  • The OP is asking for an explanation other than "it is standard"... just assume it, or offer a link only to what an answer should look like, @Dat. – amWhy Feb 09 '22 at 18:17
  • @amWhy if OP really did mean to seek an explanation for this limit, then this question must be voted to close since this already has answers elsewhere as cited. Also, it would have been really helpful if OP mentioned where they needed help. The question as is, just demands a non L'Hopital way of solving the limit, in which case the usage of such "standard" limits isperfectly fine(I believe) – DatBoi Feb 09 '22 at 19:04
  • I haven't penalized your answer in any way. Though I did act on the question. My only point is that at the level of this question, I'm not sure the OP even understands "the standard limit $\lim_{x\to 0} \frac {e^x-1}x =1$. That's why this site encourages potential answerers to seek clarification and suggest improvements, from and to the OP. For this reason, this question was not a good candidate to have answered. – amWhy Feb 09 '22 at 19:12
  • @amWhy I agree. I would have been more than glad to explain if OP read something on the cited post and had a doubt. That said, thanks for the advice, Ill take care of these for sure! :P – DatBoi Feb 09 '22 at 19:21
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As you know the graph $y=u$ is tangent to $y = e^{u}-1$ as u approaches 0, and therefore $\frac{u}{e^{u}-1}$ approaches $1$, if you substitute 2x for u,you can see that $\frac{2x}{e^{2x}-1}$ also approaches 1, so dividing it by $2$ to give $\frac{x}{e^{2x}-1}$ must give 1/2.

AmoebaProteus
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