TITLE: Limit of a Function: $ \lim_{x \to 0}\ (e^x + x)^ {\large \frac {1} {x}}$
QUESTION [7 upvotes]: What is the limit of the following function which consists of an exponential and algebraic expression? $$ \lim_{x \to 0}\ (e^x + x)^ {\large \frac {1} {x}}\;\;?$$

REPLY [2 votes]: \begin{eqnarray}
\lim_{x \to 0} (e^x+x)^{\frac{1}{x}} &=& \lim_{x \to 0} e^{\frac{1}{x} \ln (e^x + x)}\\
&=& \lim_{x \to 0} e^{\frac{1}{x} \ln (e^x(1 + xe^{-1}))} \\
&=& \lim_{x \to 0} e^{\frac{1}{x} \ln e^x} e^{\frac{1}{x} \ln (1 + xe^{-x})}\\
&=& e \lim_{x \to 0} e^{\frac{1}{x} \ln (1 + xe^{-x})}
\end{eqnarray}
Since $y \mapsto \ln(1+y)$ is differentiable at zero, for any $\epsilon>0$, there is a $\delta>0$ such that if $|y|<\delta$, then $|\ln(1+y)-y|\leq \epsilon |y|$. This gives $|\ln (1 + xe^{-x})-x e^{-x}| \leq \epsilon |x| e^{-x}$, or equivalently, $|\frac{1}{x}\ln (1 + xe^{-x})-e^{-x}| \leq \epsilon  e^{-x}$, from which it follows that $\lim_{x \to 0} \frac{1}{x}\ln (1 + xe^{-x}) = 1$, from which we obtain $\lim_{x \to 0} (e^x+x)^{\frac{1}{x}} = e^2$.