Q. 71 - Calculus

Question

Calculate each limit in Exercises 35–80. 

$\underset{x\to 0}{lim}\frac{{\sin }^{2}3x}{x^3-x}$

Step-by-Step Solution

Answer

The limit is 0

Step 1. Given information

The given expression is $\underset{x\to 0}{lim}\frac{{\sin }^{2}3x}{x^3-x}$

Step 2. Calculation

The limit is calculated as below, 

$$\begin{gathered} \underset{x\to 0}{lim}\frac{{\sin }^{2}3x}{x^3-x}=\underset{x\to 0}{lim}\frac{1-{\cos }^{2}3x}{x(x^2-1)} \\ =\underset{x\to 0}{lim}\frac{(1-\cos 3x)(1+\cos 3x)}{x(x^2-1)} \\ =\underset{x\to 0}{lim}\frac{1-\cos 3x}{x}\underset{x\to 0}{lim}\frac{1+\cos 3x}{x^2-1} \\ =\left[\underset{x\to 0}{lim}\frac{1-\cos 3x}{3x}3\right]\underset{x\to 0}{lim}\frac{1+\cos 3x}{x^2-1} \\ =3\left(0\right)\left[\frac{1+\cos 3\left(0\right)}{0-1}\right] \\ =0 \end{gathered}$$