How to prove this? $ \lim_{x \to 0}\frac{e^x-1}{x}=1 $

Question

Any idea how do I prove the following?

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

Thanks

Answer 1

Hint: let $f(x)=e^x$. What is $f'(0)$?

Answer 2

From l'Hospital's rule:

$$(e^{x}-1)'=e^{x}$$

$$x'=1$$

So you get: $\frac{e^{x}}{1} \longrightarrow1$, when $x \longrightarrow0$, because $e^0=1$.

Answer 3

Using the series expansion of $e^x$:

$$ \lim_{x \to 0}\frac{x + \frac{x^2}{2} + \frac{x^3}{6} + ~...}{x} = \lim_{x \to 0} 1 + \frac{x}{2} + \frac{x^2}{6} + ~... = 1 $$