How do you find the limit $(e^x-1)/x$ as $x->0$ ?

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Answer 1 · 2016-12-08T12:57:40

$lim_"x->0" (e^x-1)/x =1$

Since the function $(e^x-1)/x$ is continuous and undefined $(0/0)$ at $x=0$ we can apply L'Hôpital's rule for the limit as $x->0$

Thus: $lim_"x->0" (e^x-1)/x = lim_"x->0" (e^x-0)/1 =e^0=1$

How do you find the limit (e^x-1)/x(e^x-1)/x as x->0x->0?