Chapter 7: Transcendental Functions - Section 7.5 - Indeterminate Forms and L'Hopital's Rule - Exercises 7.5 - Page 409: 2

$5$

Consider: $\lim\limits_{x \to 0}f(x)=\lim\limits_{x \to 0}\dfrac{\sin 5x}{x}$ Need to check that the limit has an indeterminate form. Thus, $f(0)=\dfrac{\sin 0}{0}=\dfrac{0}{0}$ Now, apply L-Hospital's rule such as: $\lim\limits_{a \to b}f(x)=\lim\limits_{a \to b}\dfrac{g'(x)}{h'(x)}$ This implies: $\lim\limits_{x \to 0}\dfrac{5 \cos 5x}{1}=\dfrac{5 \cos 0}{1}=5$