Answer
$$ + \infty $$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\sin x}}{{{x^2}}} \cr
& {\text{Evaluate using theorem 1}}{\text{.2}}{\text{.2}} \cr
& \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\sin x}}{{{x^2}}} = \frac{{\sin \left( 0 \right)}}{{{0^2}}} = \frac{0}{0} \cr
& {\text{The numerator and denominator have a limit of 0}}{\text{, so the limit is }} \cr
& {\text{an indeterminate form of type 0/0}}{\text{. Use L'Hopital's rule}} \cr
& \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\sin x}}{{{x^2}}} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{d/dx\left( {\sin x} \right)}}{{d/dx\left( {{x^2}} \right)}} \cr
& \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\sin x}}{{{x^2}}} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\cos x}}{{2x}} \cr
& {\text{evaluating}} \cr
& = \frac{{\cos \left( 0 \right)}}{{2\left( {{0^ + }} \right)}} = \frac{1}{{{0^ + }}} \cr
& {\text{We find that 1 is positive and the denominator tends to a positive}} \cr
& {\text{number}}{\text{, so }} \cr
& \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\sin x}}{{{x^2}}} = + \infty \cr} $$
