Answer
$e$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to \infty } {\left( {{e^x} + 10x} \right)^{1/x}} \cr
& {\text{Evaluating the limit when }}x \to \infty \cr
& \mathop {\lim }\limits_{x \to \infty } {\left( {{e^x} + 10x} \right)^{1/x}} = {\left( {{{\mathop {\lim }\limits_{x \to \infty } }e^{x}} + 10\mathop {\lim }\limits_{x \to \infty } x} \right)^{\mathop {\lim }\limits_{x \to \infty } \frac{1}{x}}} = {\infty ^0} \cr
& {\text{This limit has the form }}{\infty ^0}{\text{ }} \cr
& \cr
& {\text{Rewriting}}{\left( {{e^x} + 10x} \right)^{1/x}} \cr
& {\left( {{e^x} + 10x} \right)^{1/x}} = {e^{\frac{1}{x}\ln \left( {{e^x} + 10x} \right)}},{\text{ then}} \cr
& \mathop {\lim }\limits_{x \to \infty } {\left( {{e^x} + 10x} \right)^{1/x}} = \mathop {\lim }\limits_{x \to \infty } {e^{\frac{1}{x}\ln \left( {{e^x} + 10x} \right)}} = {e^{\mathop {\lim }\limits_{x \to \infty } \frac{1}{x}\ln \left( {{e^x} + 10x} \right)}} \cr
& {\text{The first step is to evaluate }} \cr
& L = \mathop {\lim }\limits_{x \to \infty } \frac{1}{x}\ln \left( {{e^x} + 10x} \right) = \mathop {\lim }\limits_{x \to \infty } \frac{{\ln \left( {{e^x} + 10x} \right)}}{x} = \frac{\infty }{\infty } \cr
& {\text{Using the L'Hopital's rule}} \cr
& L = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{d}{{dx}}\left[ {\ln \left( {{e^x} + 10x} \right)} \right]}}{{\frac{d}{{dx}}\left[ x \right]}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{{e^x} + 10}}{{{e^x} + 10x}}}}{1} = \mathop {\lim }\limits_{x \to \infty } \frac{{{e^x} + 10}}{{{e^x} + 10x}} \cr
& L = \mathop {\lim }\limits_{x \to \infty } \frac{{1 + \frac{{10}}{{{e^x}}}}}{{1 + \frac{{10x}}{{{e^x}}}}} = \frac{{1 + 0}}{{1 + 0}} = 1 \cr
& {\text{Therefore}}{\text{,}} \cr
& \mathop {\lim }\limits_{x \to \infty } {\left( {{e^x} + 10x} \right)^{1/x}} = {e^{\mathop {\lim }\limits_{x \to \infty } \frac{1}{x}\ln \left( {{e^x} + 10x} \right)}} = {e^L} = {e^1} \cr
& \mathop {\lim }\limits_{x \to \infty } {\left( {{e^x} + 10x} \right)^{1/x}} = e \cr} $$
