A) \[\frac{p+q}{p-q}\]
B) \[\frac{p-q}{1+pq}\]
C) \[\frac{1+pq}{1-p}\]
D) \[\frac{p+q}{1-pq}\]
Correct Answer: D
Solution :
Let \[A=\underset{x\to \infty }{\mathop{\lim }}\,{{x}^{1/x}}\] Taking log on both sides, we get \[\log A=\underset{x\to \infty }{\mathop{\lim }}\,\frac{1}{x}\log x\] \[\Rightarrow \] \[\log A=0\] \[\left( \because \underset{x\to \infty }{\mathop{\lim }}\,\frac{\log x}{{{x}^{m}}}=0\,\,\forall m>0 \right)\] \[\Rightarrow \] \[A={{e}^{0}}=1\] \[\Rightarrow \] \[\underset{x\to \infty }{\mathop{\lim }}\,{{x}^{1/x}}=1\]
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