Find limes of said function without using the L'Hospital rule 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:
$$\lim_{x\to 0} \frac{x}{e^{2x} - 1}$$
 A: 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: 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,
$$
\lim_{x\to0}\frac{\tan^2x}{(e^{2x}-1)\sin x}=\lim_{x\to0}\frac{1}{2\cos^2x}\frac{\sin x}{x}\frac{2x}{e^{2x}-1}=\frac{1}{2}
$$
A: $$\lim_{x\to 0} \frac{x}{e^{2x} - 1}=\lim_{x\to 0} \frac12\frac{2x}{e^{2x} - 1}=\frac12$$
using the standard limit $\lim_{x\to0}\frac{e^x-1}{x}=1$
A: 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.
