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

Question

I'd like help finding

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

without the use of L'Hôpital's rule.

So far I did this:

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

Basically I got nowhere. Any hints or partial solutions to help me?

By the way, the is the 1969 AP BC Multiple Choice #28.

Answer 1

One may write, as $x \to 0$, $$ \frac{e^{2x}-1}{\tan x}=2\cdot\frac{e^{2x}-1}{2x}\cdot \frac{x}{\sin x}\cdot \cos x $$ then recognize standard limits.

Answer 2

Note that $e^{2x} = 1 + 2x + o(x^2)$ as $x\to 0$ and $\tan x = x + o(x^3)$ as $x\to 0$. Hence,

$$ \lim_{x\to 0}\frac{e^{2x}-1}{\tan x} = \lim_{x\to 0}\frac{[1+2x+o(x^2)]-1}{x+o(x^3)} = \lim_{x\to 0}\frac{2x}{x} = 2. $$

Answer 3

Note that $e^{2x}-1 = 2x (1+o(1))$ where $o(1)$ denotes a term going to $0$ as $x \to 0$ (see this from the maclaurin series of $e^{2x}$).

Then, you can write the thing you're taking the limit of as $\frac{2x}{\tan x} = \frac{2 x}{\sin x} \cos x$. You know $\frac{x}{\sin x} \to 1 $ as $x \to 0$ and $\cos x \to 1$ as $x \to 0$.

Answer 4

Here's where the logic for most of the answers comes from, and how we avoid the use L'Hopital's Rule.

First, let's observe a very important limit, namely $\displaystyle \lim_{h\to0}{e^h-1\over h}$. We know that $$e = \lim_{x\to\infty}\left(1+{1\over x}\right)^{x} = \lim_{y\to0}\left(1 + y\right)^{1/y}.$$ So, using some creative algebra we have $$\begin{align}\lim_{h\to0}{e^h - 1 \over h} &= \lim_{y\to0}{y\over\ln(y+1)} \tag{Let $y = e^h - 1 \iff h = \ln(y+1)$}\\ &=\lim_{y\to0}\left({1\over{1\over y}\ln(y+1)}\right) \\ &=\lim_{y\to0}{1\over\ln[(1+y)^{1/y}]}\\&={1\over\ln(e)}\\&={1\over1} \\&= 1.\end{align}$$ Notice how the limit changes from $h\to0$ to $y\to0$ because as $x\to 0$, $y = e^h - 1 \to e^0 - 1 = 1-1=0$, hence $y\to 0$. Now, we begin the important part of proving the limit.

Using this result, it is very easy to prove that $$\begin{align}\lim_{x\to0}{e^{2x} - 1\over x} &= \lim_{x\to0}\left({2\over2}\right)\left({e^{2x}-1\over x}\right)\\&=2\lim_{x\to0}{e^{2x}-1\over2x}\\&=2(1)\\&=2.\end{align}$$

Therefore, we have that $$\begin{align}\lim_{x\to0}{e^{2x}-1\over\tan x} &=\lim_{x\to0}[(e^{2x} - 1)\cot(x)] \\ &=\lim_{x\to0}\left((e^{2x}-1)\cdot{\cos(x)\over \sin(x)}\right) \\ &=\lim_{x\to0}\left((e^{2x}-1)\cdot{2x\over2x}\cdot{\cos(x)\over \sin(x)}\right)\\&=2\lim_{x\to0}\left[{e^{2x}-1\over2x}\cdot{x\over\sin(x)}\cdot\cos(x)\right]\\&= 2\lim_{x\to0}\left[{e^{2x}-1\over2x}\cdot{1\over{\sin(x)\over x}}\cdot\cos(x)\right]\\&=2\left(1\cdot{1\over1}\cdot1\right)\\&=2.\end{align}$$