How to evaluate $\lim_{x \to 0} \frac{e^x - 1}{x}$ I can only get $$\lim_{x \to 0} \frac{e^x - 1}{x}$$ to be undefined since $\frac{0}{0}$ doesn't mean anything. How can I manipulate the expression to get a valid answer?
 A: Use the Taylor series of the exponential : 
$$e^x=\sum_0^{\infty}\frac{x^n}{n!}$$
$$\lim_{x \to 0} \frac{e^x - 1}{x}=\lim_{x \to 0} \frac{(1+x+o(x) - 1)}{x}=\lim_{x \to 0} \frac{x+o(x)}{x}=1$$
You can also recognise the definition of the derivative of a function $f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}{x}$, here $f(x)=e^x$, so :
$$\lim_{x \to 0} \frac{e^x - 1}{x}=e^0=1$$
A: This is nothing more than
$$\lim_{x\to 0} \frac{f(x) - f(0)}{x-0}$$
for $f(x) = e^x.$ Look familiar?
A: Without L'Hospital's rule we can write as follows:
Since      $ \lim _{ x\rightarrow 0 }{ { \left( 1+x \right)  }^{ \frac { 1 }{ x }  } } =e$

$${ e }^{ x }-1=t\Rightarrow x=\ln { \left( t+1 \right)  } \\ \lim _{ t\rightarrow 0 }{ \frac { t }{ \ln { \left( t+1 \right)  }  }  } =\lim _{ t\rightarrow 0 }{ \frac { 1 }{ \ln { { \left( t+1 \right)  }^{ \frac { 1 }{ t }  } }  }  } =\frac { 1 }{ \ln { e }  } =1\\ $$

A: You can do a series expansion, or you can use l'Hopital's rule.  But, If you don't know those yet.
$e^x = \lim_\limits{n\to\infty} (1+\frac xn)^n$
So, you can do a binomial expansion on that for a finite $n.$  Find your limit.  And then look at the implications as $n$ goes to infinity.
A: From the definition of $e$, we can easily obtain the following inequality
\begin{equation}
(1+x)^{\large{1\over x}}\le e\le(1-x)^{-\large{1\over x}}
\end{equation}
then applying the sandwich theorem, we obtain
\begin{equation}
1\le \lim_{x\to0}\ {e^x-1\over x}\le \lim_{x\to0}\ {1\over 1-x}=1
\end{equation}
A: You can use L'Hôpital's rule
$$\lim_{x \to 0} \frac{e^x - 1}{x}
= \lim_{x \to 0} \frac{\frac{d}{dx} (e^x - 1)}{\frac{d}{dx}x}
= \lim_{x \to 0} \frac{e^x}{1}
= 1$$
