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

$5$
Consider $f(x)=\lim\limits_{x \to 0}\dfrac{5-5 \cos x}{e^x-x -1}=\lim\limits_{x \to 0} \dfrac{a(x)}{b(x)}$ and $a(0)=0, b(0)=0$ Thus, $f(0)=\dfrac{0}{0}$ This shows an Inderminate form of the limit, so apply L-Hospital's rule: $\lim\limits_{x \to l}\dfrac{a(x)}{b(x)}=\lim\limits_{x \to l}\dfrac{a'(x)}{b'(x)}$ Thus, $\lim\limits_{x \to 0}\dfrac{5 \sin x)}{e^x-1}=\dfrac{0}{0}$ Again apply L-Hospital's rule: $\lim\limits_{x \to l}\dfrac{a(x)}{b(x)}=\lim\limits_{x \to l}\dfrac{a'(x)}{b'(x)}$ Thus, $\lim\limits_{x \to 0}\dfrac{5 \cos x}{e^x}=5$