Finding limit without using the l'Hopital theorem

Question

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).

Answer 1

\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}

Answer 2

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.

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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.

Answer 3

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 $x\to 0.$