Question

$\underset{x\to 0}{\lim }\left(\frac{e^x-1}{x}\right)$

Short answer

$\underset{x\to 0}{\lim }\left(\frac{e^x-1}{x}\right)=1$

Step-by-step solution

Step 1. Given information

$\underset{x\to 0}{\lim }\left(\frac{e^x-1}{x}\right)$

Step 2. Plug in the values

Applying L'Hopital's rule,

$\underset{x\to 0}{\lim }\left(\frac{e^x-1}{x}\right)=\underset{x\to \hspace{0.17em}0}{\lim }\left(\frac{e^x}{1}\right)$

$\underset{x\to 0}{\lim }\left(\frac{e^x-1}{x}\right)=\frac{e^0}{1}$

$\underset{x\to 0}{\lim }\left(\frac{e^x-1}{x}\right)=1$