A) \[\frac{n}{m}\]
B) \[\frac{m}{n}\]
C) \[\frac{2m}{n}\]
D) \[\frac{2n}{m}\]
Correct Answer: B
Solution :
\[\underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{m}}-1}{{{x}^{n}}-1}\] \[=\underset{x\to 1}{\mathop{\lim }}\,\,\,\frac{m{{x}^{m}}^{-1}}{n{{x}^{n}}^{-1}}\] (using L? Hospital?s rule) \[=\frac{m}{n}\]You need to login to perform this action.
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