Answer
$e^{2}$
Work Step by Step
L'Hospital's rule states that $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{A'(x)}{B'(x)}$
Here, $\ln f(x)=\dfrac{\ln (e^x+x)}{x} $
Now, $\lim\limits_{x \to 0} f(0)=\dfrac{0}{0}$
This shows an indeterminate form of the limit, so we need to use L'Hospital's rule.
$e^{\lim\limits_{x \to 0} \frac{e^x+1/e^x+x}{1}}=e^{\lim\limits_{x \to 0} \frac{e^0+1}{e^0+0}}$
Thus,
$e^{\lim\limits_{x \to 0} \frac{e^0+1}{e^0+0}}=e^{2}$
