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Any idea how do I prove the following?

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

Thanks

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    $\begingroup$ 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. $\endgroup$ – egreg Feb 5 '15 at 12:51
  • $\begingroup$ Really sorry guys... haven't done maths for many years... completely forgot about L'Hôpital's rule... Thanks for answering :) $\endgroup$ – chengcj Feb 5 '15 at 13:27
  • $\begingroup$ 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. $\endgroup$ – chengcj May 28 '15 at 14:23
  • $\begingroup$ math.stackexchange.com/questions/1828962/… $\endgroup$ – user301988 Jul 4 '16 at 0:24
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Using the series expansion of $e^x$:

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

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  • $\begingroup$ 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. $\endgroup$ – Jason Feb 5 '15 at 22:44
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From l'Hospital's rule:

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

$$x'=1$$

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

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    $\begingroup$ 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... $\endgroup$ – Hans Lundmark Feb 5 '15 at 13:09
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Hint: let $f(x)=e^x$. What is $f'(0)$?

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