Chapter 7: Transcendental Functions - Section 7.5 - Indeterminate Forms and L'Hopital's Rule - Exercises 7.5 - Page 410: 58

$e^{2}$

Here, we have $\ln f(x)=\dfrac{\ln (e^x+x)}{x} $ But $\lim\limits_{x \to 0} f(0)=\dfrac{0}{0}$ This shows an indeterminate form of limit, thus we will apply L-Hospital's rule such as: $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$ $e^{\lim\limits_{x \to 0} \dfrac{e^x+1/e^x+x}{1}}=e^{\dfrac{e^0+1}{e^0+0}}$ or, $e^{\frac{1+1}{1}}=e^{2}$