Question

Find the limit.

48. \(\mathop {lim}\limits_{x \to 0} \frac{{si{n^2}3x}}{x}\)

Short answer

The value of the limit is 0.

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:

\(\begin{array}{c}\mathop {\lim }\limits_{x \to \infty } \frac{{{{\sin }^2}3x}}{x}{\rm{ }} = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sin 3x} \right)\left( {\sin 3x} \right)}}{x}\\ = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin 3x}}{x}} \right)\mathop {\lim }\limits_{x \to 0} \left( {\sin 3x} \right)\\ = \lim \left( {\frac{{\sin 3x}}{{3x}}} \right) \cdot 3 \times \sin \left( 0 \right)\\ = 1 \times 3 \times 0\\ = 0\end{array}\)

Thus, the value of the limit is 0.