## Calculus (3rd Edition)

The limit $\lim _{x \rightarrow 0} \frac{\sin x}{x^{2}}$ does not exist.
We have $$\lim _{x \rightarrow 0} \frac{\sin x}{x^{2}}=\lim _{x \rightarrow 0} \frac{\sin x}{x} \lim _{x \rightarrow 0} \frac{1}{x}= \lim _{x \rightarrow 0} \frac{1}{x}.$$ Where we used the fact that $\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$. Now, the limit $\lim _{x \rightarrow 0} \frac{1}{x}$ does not exist. This is because the left and right hand limits are not the same: $$\lim _{x \rightarrow 0^+} \frac{1}{x}=\infty,\quad \lim _{x \rightarrow 0^-} \frac{1}{x}=-\infty.$$ Hence, the limit $\lim _{x \rightarrow 0} \frac{\sin x}{x^{2}}$ does not exist.