Chapter 1 - Limits and Continuity - 1.6 Continuity of Trigonometric Functions - Exercises Set 1.6 - Page 106: 23

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Given $$\lim_{x\to 0 }\frac{\sin x^2}{x}$$ Then \begin{align*} \lim_{x\to 0 }\frac{\sin x^2}{x}&=\lim_{x\to 0 }x\frac{\sin x^2}{x^2}\\ &=\left(\lim_{x\to 0 }x\right)\left(\lim_{x\to 0 }\frac{\sin x}{ x } \right)^2 \\ &=(0)(1)\\ &=0 \end{align*}