A) \[m/n\]
B) \[{{m}^{2}}/{{n}^{2}}\]
C) \[0\]
D) \[n/m\]
Correct Answer: B
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{1-\cos mx}{1-\cos nx}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{1-\left( 1-\frac{{{(mx)}^{2}}}{2!}+\frac{{{(mx)}^{4}}}{4!}-... \right)}{1-\left( 1-\frac{{{(nx)}^{2}}}{2!}+\frac{{{(nx)}^{4}}}{4!}+... \right)}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{2}}\left( \frac{{{m}^{2}}}{2!}+\frac{{{m}^{4}}{{x}^{2}}}{4!}+... \right)}{{{x}^{2}}\left( \frac{{{n}^{2}}}{2!}-\frac{{{n}^{4}}{{x}^{2}}}{4!}+... \right)}\] \[=\frac{{{m}^{2}}}{{{n}^{2}}}\] Alternative Solution: \[\underset{x\to 0}{\mathop{\lim }}\,\frac{1-\cos mx}{1-\cos nx}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{\sin }^{2}}\left( \frac{mx}{2} \right)}{2{{\sin }^{2}}\left( \frac{nx}{2} \right)}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{\frac{\sin \left( \frac{mx}{2} \right)\times \left( \frac{{{m}^{2}}{{x}^{2}}}{4} \right)}{\left( \frac{{{n}^{2}}{{x}^{2}}}{4} \right)}}{\frac{{{\sin }^{2}}\left( \frac{nx}{2} \right)\times \left( \frac{{{n}^{2}}{{x}^{2}}}{4} \right)}{\left( \frac{{{n}^{2}}{{x}^{2}}}{4} \right)}}\] \[=\frac{{{m}^{2}}}{{{n}^{2}}}\]
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