A) \[\frac{5}{3}\]
B) \[\frac{3}{7}\]
C) \[\frac{4}{7}\]
D) \[\frac{-3}{5}\]
Correct Answer: A
Solution :
\[\underset{x\to -1}{\mathop{\lim }}\,\frac{1+{{x}^{\frac{1}{2}}}}{1+{{x}^{\frac{1}{5}}}}=\underset{x\to -1}{\mathop{\lim }}\,\frac{a+\frac{1}{3}{{x}^{-\frac{2}{3}}}}{0+\frac{1}{5}{{x}^{-\frac{4}{5}}}}\] (using L? Hospital?s rule) \[=\underset{x\to -1}{\mathop{\lim }}\,\frac{5}{3}{{x}^{\frac{4}{5}-\frac{2}{3}}}\] \[=\underset{x\to -1}{\mathop{\lim }}\,\frac{5}{3}{{(x)}^{\frac{2}{15}}}=\frac{5}{3}{{[{{(-1)}^{2}}]}^{\frac{1}{15}}}=\frac{5}{3}\]
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