Question

45-60: Find the limit.

54. \(\mathop {\lim }\limits_{x \to 0} \frac{{\sin 3x\sin 5x}}{{{x^2}}}\)

Short answer

The required value is \(\mathop {\lim }\limits_{x \to 0} \frac{{\sin 3x\sin 5x}}{{{x^2}}} = 15\).

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 \(\mathop {\lim }\limits_{x \to 0} \frac{{\sin 3x\sin 5x}}{{{x^2}}}\).

Manipulate and solve the given limit as follows:

\(\begin{aligned}\mathop {\lim }\limits_{x \to 0} \frac{{\sin 3x\sin 5x}}{{{x^2}}} &= \mathop {\lim }\limits_{x \to 0} \frac{{\sin 3x\sin 5x}}{{{x^2}}} \cdot \frac{5}{5} \cdot \frac{3}{3}\\ &= \mathop {\lim }\limits_{x \to 0} \frac{{\sin 3x\sin 5x}}{{x \cdot x}} \cdot \frac{5}{5} \cdot \frac{3}{3}\\ &= \mathop {\lim }\limits_{x \to 0} \frac{{\sin 3x}}{{3x}} \cdot \frac{{\sin 5x}}{{5x}} \cdot 5 \cdot 3\\ &= \mathop {\lim }\limits_{x \to 0} \frac{{\sin 3x}}{{3x}} \cdot \frac{{\sin 5x}}{{5x}} \cdot 15\end{aligned}\)

Solve the above limit further as follows:

\(\begin{aligned}\mathop {\lim }\limits_{x \to 0} \frac{{\sin 3x\sin 5x}}{{{x^2}}} &= \mathop {\lim }\limits_{x \to 0} \frac{{\sin 3x}}{{3x}} \cdot \mathop {\lim }\limits_{x \to 0} \frac{{\sin 5x}}{{5x}} \cdot \mathop {\lim }\limits_{x \to 0} 15\\ &= \mathop {\lim }\limits_{3x \to 0} \frac{{\sin 3x}}{{3x}} \cdot \mathop {\lim }\limits_{5x \to 0} \frac{{\sin 5x}}{{5x}} \cdot \mathop {\lim }\limits_{x \to 0} 15\\ &= 1 \cdot 1 \cdot \mathop {\lim }\limits_{x \to 0} 15\\ &= 15\end{aligned}\)

Hence, \(\mathop {\lim }\limits_{x \to 0} \frac{{\sin 3x\sin 5x}}{{{x^2}}} = 15\).