# Chapter 2 - Limits - 2.2 Limits: A Numerical and Graphical Approach - Exercises - Page 54: 24

The limit $\lim _{x \rightarrow 0} \frac{\sin x}{x^{2}}$ does not exist.

#### Work Step by Step

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.

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