Question

45-60: Find the limit.

58. \(\mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {{x^2}} \right)}}{x}\)

Short answer

The required value is \(\mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {{x^2}} \right)}}{x} = 0\).

Step-by-step solution

Special Trigonometric limits

The special trigonometric for sine function is given below:

\(\mathop {\lim }\limits_{\theta \to 0} \frac{{\sin \theta }}{\theta } = 1\)

Limit evaluation

The required limit is\(\mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {{x^2}} \right)}}{x}\).

Manipulate and solve the given limit as follows:

\(\begin{aligned}\mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {{x^2}} \right)}}{x} &= \mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {{x^2}} \right)}}{x} \cdot \frac{x}{x}\\ &= \mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {{x^2}} \right)}}{{{x^2}}} \cdot x\\ &= \mathop {\lim }\limits_{{x^2} \to 0} \frac{{\sin \left( {{x^2}} \right)}}{{{x^2}}} \cdot \mathop {\lim }\limits_{x \to 0} x\\ &= 1 \cdot \mathop {\lim }\limits_{x \to 0} x\\ &= 0\end{aligned}\)

Hence, \(\mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {{x^2}} \right)}}{x} = 0\).