A) 0
B) \[\frac{1}{2}\]
C) \[\frac{1}{p}-\frac{1}{p-1}\]
D) \[{{e}^{4}}\]
Correct Answer: B
Solution :
\[\underset{x\to \infty }{\mathop{\lim }}\,\,\left[ \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x} \right]=\underset{x\to \infty }{\mathop{\lim }}\,\frac{x+\sqrt{x+\sqrt{x}}-x}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}\] \[=\underset{x\to \infty }{\mathop{\lim }}\,\frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}=\underset{x\to \infty }{\mathop{\lim }}\,\,\frac{\sqrt{1+{{x}^{-1/2}}}}{\sqrt{1+\sqrt{{{x}^{-1}}+{{x}^{-3/2}}}}+1}=\frac{1}{2}\].
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