Answer
$1$
Work Step by Step
Here, we have $\lim\limits_{x \to 0^{+}} f(0)=\dfrac{0}{0}$
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)}$$
\lim\limits_{x \to 0^{+}} \dfrac{e^x/e^x-1}{1/x}=\lim\limits_{x \to 0^{+}} \dfrac{xe^x}{e^x-1}=\dfrac{0}{0}$
Now, again apply L-Hospital's rule.
$\lim\limits_{x \to 0^{+}} \dfrac{e^x+xe^x}{e^x}=1$
