Question

Find the limit.

49. \(\mathop {\lim }\limits_{x \to 0} \frac{{\sin x - \sin x\cos x}}{{{x^2}}}\)

Short answer

The value of the limit is \(\infty \).

Step-by-step solution

Use the property of limit

A function is continuous at a point “a” if \(\mathop {\lim }\limits_{x \to a} f\left( x \right) = f\left( a \right)\).

Evaluate the given limit

Evaluate the given limit as follows:

\(\begin{aligned}\mathop {\lim }\limits_{x \to \infty } \frac{{\sin x - \sin x\cos x}}{{{x^2}}}{\rm{ }} &= \mathop {\lim }\limits_{x \to 0} \frac{{\sin x - \sin x.\cos x}}{{{x^2}}}\\ &= \mathop {\lim }\limits_{x \to \infty } \frac{{\sin x\left( {1 - \cos x} \right)}}{{{x^2}}}\\ &= \mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} \cdot \mathop {\lim }\limits_{x \to 0} \frac{{2{{\sin }^2}\frac{x}{2}}}{x}\\ &= 2 \times 1 \times \mathop {\lim }\limits_{x \to 0} {\left( {\frac{{\sin \left( {\frac{x}{2}} \right)}}{{\left( {\frac{x}{2}} \right)}}} \right)^2}{\left( {\frac{x}{2}} \right)^2} \times \frac{1}{x}\\ &= 2 \times 1 \times 1 \times \mathop {\lim }\limits_{x \to 0} \frac{x}{4}\\ &= 2 \times 0\\ &= 0\end{aligned}\)

Thus, the value of the limit is\(\infty \).