# Finding limit without using the l'Hopital theorem

I was trying to find two limits which are trivial using l'Hopital, but I can't seem to find a way to do it without it. For example : $$f(x)= \frac{e^{2\sin(x)} -1}{x}$$

I tried to do it using the fact I know $$\lim \frac{e^x-1}{x} = 1$$ as $$x \to 0$$. Then, as $$2\sin(x)$$ tends to $$0$$ as well as $$x \to 0$$, I first tried to prove the limit of $$f(x)$$ would be $$1$$, but I know it is $$2$$ (using l'hôpital).

\begin{align} \lim_{x \to 0}\frac{e^{2\sin x} - 1}{x} &=\lim_{x \to 0}\frac{2\sin x}{x}\frac{e^{2\sin x} - 1}{2 \sin (x)} \\ &= 2\lim_{x \to 0} \frac{\sin x}{x}\cdot \lim_{x \to 0}\frac{e^{2\sin x} - 1}{2 \sin (x)}\\ &= 2\lim_{x \to 0} \frac{\sin x}{x}\cdot \lim_{y \to 0}\frac{e^{y} - 1}{y}\\ &= 2 \end{align}

You are using L'Hopital's rule to find a derivative (at $$0$$). That is, you are using a non-trivial theorem asking to compute 2 derivatives (that of $$f$$ and that of $$g$$, applying L'Hopital to something like $$\frac{f}{g}$$) on a neighborhood around $$0$$, in order to find one (that of $$f$$ only) at the single point $$0$$.

This is overkill.

Define the function $$f$$ by $$f(x) = e^{2\sin x}$$ for $$x\in\mathbb{R}$$. Note that $$f(0) = 1$$.

Then, $$\lim_{x\to 0}\frac{e^{2\sin x}-1}{x} = \lim_{x\to 0}\frac{f(x)-f(0)}{x-0}$$ and this is the definition of $$f'(0)$$ (if it exists). But it exists: by composition, $$f$$ is differentiable on $$\mathbb{R}$$, and $$f'(x) = (2\cos x)e^{2\sin x}$$, so $$f'(0) = 2$$.

No need for L'Hopital.

As for why your attempt failed: let $$u(x)=2\sin x$$. Then indeed $$u(x)\to 0$$ as $$x\to 0$$. That means $$\lim_{x\to 0} \frac{e^{u(x)}}{u(x)}=1$$, though, not $$\lim_{x\to 0} \frac{e^{u(x)}}{x}=1$$. So you need to rewrite $$\frac{e^{u(x)}}{u(x)} = \frac{e^{u(x)}}{x}\cdot \frac{x}{u(x)} \xrightarrow[x\to0]{} 1\cdot \lim_{x\to0}\frac{x}{u(x)}$$ which will give you the right answer.

For another approach, note that $$e^{2\sin x}=1+2\sin x+o(\sin^2x)$$ so $$\frac{e^{2\sin(x)} -1}{x}=\frac{2\sin x+o(\sin^2x)}{x}\to 2$$ as $$x\to 0.$$