Q104 / 170 · Manipal (MET) 2017 · Mathematics

$\lim _{x \rightarrow 0} \frac{1-\cos m x}{1-\cos n x}$ is equal to

  • A
    $\frac{m}{n}$
  • Correct answer
    B
    $\frac{m^2}{n^2}$
  • C
    0
  • D
    $\frac{n^2}{m^2}$

Solution

$\begin{aligned} & \lim _{x \rightarrow 0}\left(\frac{1-\cos m x}{1-\cos n x}\right) \\ = & \lim _{x \rightarrow 0} \frac{2 \sin ^2(m x / 2)}{(m x / 2)^2} \cdot \frac{(n x / 2)^2}{2 \sin ^2(n x / 2)} \cdot \frac{m^2}{n^2}=\frac{m^2}{n^2}\end{aligned}$

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

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