MCQ
$\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos mx}}{{1 - \cos nx}} = $
  • A
    $m/n$
  • B
    $n/m$
  • $\frac{{{m^2}}}{{{n^2}}}$
  • D
    $\frac{{{n^2}}}{{{m^2}}}$

Answer

Correct option: C.
$\frac{{{m^2}}}{{{n^2}}}$
c
(c) $\mathop {\lim }\limits_{x \to 0} \,\frac{{1 - \cos mx}}{{1 - \cos \,nx}} = \mathop {\lim }\limits_{x \to 0} \,\,\left\{ {\frac{{2\,{{\sin }^2}{\textstyle{{mx} \over 2}}}}{{2\,{{\sin }^2}{\textstyle{{nx} \over 2}}}}} \right\}$

$ = \mathop {\lim }\limits_{x \to 0} \,\left[ {{{\left\{ {\frac{{\sin {\textstyle{{mx} \over 2}}}}{{{\textstyle{{mx} \over 2}}}}} \right\}}^2}\,\,\,\frac{{{m^2}{x^2}}}{4}.\frac{1}{{{{\left\{ {\frac{{\sin {\textstyle{{nx} \over 2}}}}{{{\textstyle{{nx} \over 2}}}}} \right\}}^2}}}.\frac{4}{{{n^2}{x^2}}}} \right]$

$ = \frac{{{m^2}}}{{{n^2}}} \times 1 = \frac{{{m^2}}}{{{n^2}}}$.

Aliter : Apply $L$-Hospital’s rule,

$\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos mx}}{{1 - \cos nx}} = \mathop {\lim }\limits_{x \to 0} \frac{{m\sin mx}}{{n\sin nx}} $

$= \mathop {\lim }\limits_{x \to 0} \frac{{{m^2}\cos mx}}{{{n^2}\cos nx}} = \frac{{{m^2}}}{{{n^2}}}.$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free