Q. 72 - Calculus

Question

Calculate each limit in Exercises 35–80. 

$\underset{x\to 0}{lim}\left(\frac{\sin \left(3x^2\right)}{x^3-x}\right)$

Step-by-Step Solution

Answer

The limit is 0

Step 1. Given information

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

Step 2. Calculation

The limit is calculated as below, 

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