Answer
Please see below.
$$\begin{array}{|c|c|c|c|c|c|c|c|} \hline
x & -0.1 & -0.01 & - 0.001 & 0 & 0.001 & 0.01 & 0.1 \\ \hline
\frac{\sin x^2}{x} & -0.099998333 & -0.01 & -0.001 & \text{undefined} & 0.001 & 0.01 & 0.099998333 \\ \hline
\end{array}$$
Work Step by Step
Looking at the graph, we find that when $x$ approaches $0$ from the left and right, the function approaches $0$. So, we can conclude that$$\lim_{x \to 0}\frac{\sin x^2}{x}=0 \, .$$
The table confirms our result.
Now, we find the limit analytically. Since by direct substitution we get the indeterminate form $\frac{0}{0}$, we multiply the numerator and denominator by $x$ (We can do this since $x \neq 0$). So, we have$$\lim_{x \to 0}\frac{\sin x^2}{x}=\lim_{x \to 0}\left ( \vphantom{\frac{\sin x^2}{x^2}} x \right ) \left (\frac{\sin x^2}{x^2} \right )= (0)(1)=0$$(In finding the limit we have used Thereom 1.9 (a), $\lim_{x \to 0} \frac{\sin x}{x}=1$, also $u=x^2$).
