A) \[\frac{44}{91}\]
B) \[\frac{45}{91}\]
C) \[\frac{45}{89}\]
D) \[\frac{45}{89}\]
Correct Answer: B
Solution :
Let \[L=\underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{\frac{1}{13}}}-{{x}^{\frac{1}{7}}}}{{{x}^{\frac{1}{5}}}-{{x}^{\frac{1}{3}}}}\] \[\left( \frac{0}{0}\text{form} \right)\] Apply L? Hospital?s rule, we get \[L=\underset{x\to 1}{\mathop{\lim }}\,\frac{\frac{1}{13}{{x}^{-1+\frac{1}{13}}}-\frac{1}{7}{{x}^{-1+\frac{1}{7}}}}{\frac{1}{5}{{x}^{\frac{1}{5}-1}}-\frac{1}{3}{{x}^{\frac{1}{3}-1}}}\] \[L=\underset{x\to 1}{\mathop{\lim }}\,\frac{\frac{1}{13}-\frac{1}{7}}{\frac{1}{5}-\frac{1}{3}}=\frac{45}{91}\]
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