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

Any idea how do I prove the following?

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

Thanks

• This is a “chicken and egg” problem: we can't know what hint to give if you don't say how the exponential function has been defined for you. – egreg Feb 5 '15 at 12:51
• Really sorry guys... haven't done maths for many years... completely forgot about L'Hôpital's rule... Thanks for answering :) – chengcj Feb 5 '15 at 13:27
• By the way, I'm new to math stack exchange. How did you guys find out that this question was duplicated? Is there a clever way to search for equations / formulae in math stack exchange? Thanks. – chengcj May 28 '15 at 14:23
• math.stackexchange.com/questions/1828962/… – user301988 Jul 4 '16 at 0:24

Using the series expansion of $e^x$:

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

• I would be careful about taking limits of an infinite series. It's probably better to use Taylor's theorem to approximate the remainder so you end up with a finite sum. – Jason Feb 5 '15 at 22:44

From l'Hospital's rule:

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

$$x'=1$$

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

• Watch out for circular reasoning here! The limit in question is usually used when proving that $(e^x)'=e^x$, and in that case you can't use derivatives to show what the limit is... – Hans Lundmark Feb 5 '15 at 13:09

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