Question

What is $\underset{x\to 0}{lim}\frac{\sin 2x+4x}{2x+\sin 4x}$ equal to?

• A. $0$
• B. $\frac{1}{2}$
• C. $1$
• D. $2$

Answer 1

Answer: (C)

$1$

We know, $\underset{x\to 0}{lim}\frac{\sin x}{x}=1$ .... $\left(i\right)$

Then, $\underset{x\to 0}{lim}\frac{\sin 2x+4x}{2x+\sin 4x}$

$=\underset{x\to 0}{lim}\frac{\frac{\sin 2x}{x}+4}{2+\frac{\sin 4x}{x}}$

$=\underset{x\to 0}{lim}\frac{2\frac{\sin 2x}{2x}+4}{2+4\frac{\sin 4x}{4x}}$

$=\frac{2+4}{2+4}=1$ ...... Using $\left(i\right)$

Hence, C is correct.