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