Question

$\underset{x\to 0}{\lim }\frac{e^{2\tan x}-1}{x}$ ​​is equal to _____________________.

Answer 1

$$\begin{gathered} \begin{array}{c}\lim \\ x\to 0\end{array}\frac{e^{2\tan x}-1}{x} \\ \mathrm{i}.\mathrm{e}.\begin{array}{c}\lim \\ x\to 0\end{array}\frac{e^{2\tan x}-1}{2\tan x}\times \frac{2\tan x}{x} \\ \mathrm{i}.\mathrm{e}.\begin{array}{c}\lim \\ x\to 0\end{array}\left(\frac{e^{2\tan x}-1}{2\tan x}\right)\times 2\left(\frac{\tan x}{x}\right) \\ \mathrm{i}.\mathrm{e}.\left(\begin{array}{c}\lim \\ x\to 0\end{array}\frac{e^{2\tan x}-1}{2\tan x}\right)\times 2\left(\begin{array}{c}\lim \\ x\to 0\end{array}\frac{\tan x}{x}\right) \\ \mathrm{i}.\mathrm{e}.1\times 2\times 1=2 \\ \mathrm{i}.\mathrm{e}.\begin{array}{c}\lim \\ x\to 0\end{array}\frac{e^{2\tan x}-1}{x}=2 \end{gathered}$$