Question

In Problems 53-56, use the properties of limits and the facts that

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{\sin x}{x}=1 \\ \underset{x\to 0}{\lim }\frac{\cos x-1}{x}=0 \\ \underset{x\to 0}{\lim }\sin x=0 \\ \underset{x\to 0}{\lim }\cos x=1 \end{gathered}$$

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

Short answer

The limit is$2$

Step-by-step solution

Given information

Given function$\underset{x\to 0}{\lim }\frac{\sin \left(2x\right)}{x}$

Evaluating the limit by using double angle formula

Using double angle formula, we get

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{\sin \left(2x\right)}{x}=\underset{x\to 0}{\lim }\frac{2\sin x\cos x}{x} \\ \underset{x\to 0}{\lim }\frac{\sin \left(2x\right)}{x}=2\underset{x\to 0}{\lim }\frac{\sin x}{x}.\underset{x\to 0}{\lim }\cos x \\ \underset{x\to 0}{\lim }\frac{\sin \left(2x\right)}{x}=2.1.1 \\ \underset{x\to 0}{\lim }\frac{\sin \left(2x\right)}{x}=2 \end{gathered}$$