Question

45-60: Find the limit.

53. \(\mathop {\lim }\limits_{x \to 0} \frac{{\sin 3x}}{{5{x^3} - 4x}}\)

Short answer

\(\mathop {\lim }\limits_{x \to 0} \frac{{\sin 3x}}{{5{x^3} - 4x}} = - \frac{3}{4}\)

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 \(\mathop {\lim }\limits_{x \to 0} \frac{{\sin 3x}}{{5{x^3} - 4x}}\).

Manipulate and solve the given limit as follows:

\(\begin{aligned}\mathop {\lim }\limits_{x \to 0} \frac{{\sin 3x}}{{5{x^3} - 4x}} &= \mathop {\lim }\limits_{x \to 0} \frac{{\sin 3x}}{{x\left( {5{x^2} - 4} \right)}}\\ &= \mathop {\lim }\limits_{x \to 0} \frac{{\sin 3x}}{{x\left( {5{x^2} - 4} \right)}} \cdot \frac{3}{3}\\ &= \mathop {\lim }\limits_{x \to 0} \frac{{\sin 3x}}{{3x}} \cdot \frac{3}{{\left( {5{x^2} - 4} \right)}}\\ &= \mathop {\lim }\limits_{x \to 0} \frac{{\sin 3x}}{{3x}} \cdot \mathop {\lim }\limits_{x \to 0} \frac{3}{{\left( {5{x^2} - 4} \right)}}\end{aligned}\)

Solve the above limit further as follows:

\(\begin{aligned}\mathop {\lim }\limits_{x \to 0} \frac{{\sin 3x}}{{5{x^3} - 4x}} &= \mathop {\lim }\limits_{3x \to 0} \frac{{\sin 3x}}{{3x}} \cdot \mathop {\lim }\limits_{x \to 0} \frac{3}{{\left( {5{x^2} - 4} \right)}}\\ &= 1 \cdot \mathop {\lim }\limits_{x \to 0} \frac{3}{{\left( {5{x^2} - 4} \right)}}\\ &= \frac{3}{{5{{\left( 0 \right)}^2} - 4}}\\ &= \frac{3}{{ - 4}}\\ &= - \frac{3}{4}\end{aligned}\)

Hence, \(\mathop {\lim }\limits_{x \to 0} \frac{{\sin 3x}}{{5{x^3} - 4x}} = - \frac{3}{4}\).