## Thomas' Calculus 13th Edition

$1$
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. $\lim\limits_{x \to 0^{+}} \dfrac{e^x+xe^x}{e^x}=1$