A) \[m/n\]
B) \[n/m\]
C) \[\frac{{{m}^{2}}}{{{n}^{2}}}\]
D) \[\frac{{{n}^{2}}}{{{m}^{2}}}\]
Correct Answer: C
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\,\frac{1-\cos mx}{1-\cos \,nx}=\underset{x\to 0}{\mathop{\lim }}\,\,\,\left\{ \frac{2\,{{\sin }^{2}}\tfrac{mx}{2}}{2\,{{\sin }^{2}}\tfrac{nx}{2}} \right\}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\,\left[ {{\left\{ \frac{\sin \tfrac{mx}{2}}{\tfrac{mx}{2}} \right\}}^{2}}\,\,\,\frac{{{m}^{2}}{{x}^{2}}}{4}.\frac{1}{{{\left\{ \frac{\sin \tfrac{nx}{2}}{\tfrac{nx}{2}} \right\}}^{2}}}.\frac{4}{{{n}^{2}}{{x}^{2}}} \right]\] \[=\frac{{{m}^{2}}}{{{n}^{2}}}\times 1=\frac{{{m}^{2}}}{{{n}^{2}}}\]. Aliter : Apply L-Hospital?s rule, \[\underset{x\to 0}{\mathop{\lim }}\,\frac{1-\cos mx}{1-\cos nx}=\underset{x\to 0}{\mathop{\lim }}\,\frac{m\sin mx}{n\sin nx}=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{m}^{2}}\cos mx}{{{n}^{2}}\cos nx}=\frac{{{m}^{2}}}{{{n}^{2}}}.\]
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