## University Calculus: Early Transcendentals (3rd Edition)

$1$
L'Hospital's rule states that $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{A'(x)}{B'(x)}$ Here, $\lim\limits_{x \to 0^{+}} f(0)=\dfrac{0}{0}$ This shows an indeterminate form of a limit, so we need to use L'Hospital's rule. $\lim\limits_{x \to 0^{+}} \dfrac{e^x/e^x-1}{1/x}=\lim\limits_{x \to 0^{+}} \dfrac{xe^x}{e^x-1}=\dfrac{0}{0}$ Again apply L'Hospital's rule. $\lim\limits_{x \to 0^{+}} \dfrac{e^x+xe^x}{e^x}=\dfrac{e^0+(0)}{e^(0)}=1$