Q213 / 270 · NDA 2013 (Phase 2) · Mathematics

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

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

Solution

$\begin{aligned} & \lim _{x \rightarrow 0} \frac{\sin 2 x+4 x}{2 x+\sin 4 x} \\ & =\lim _{x \rightarrow 0} \frac{\frac{2 \cdot \sin 2 x}{2 x}+\frac{4 x}{x}}{\frac{2 x}{x}+\frac{4 \cdot \sin 4 x}{4 x}} \\ & =\frac{2+4}{2+4}=\frac{6}{6}=1\end{aligned}$

Hence, the correct answer is option: 'C'.

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