A) 2
B) 4
C) 1
D) ½
Correct Answer: A
Solution :
\[y=\underset{x\to 1}{\mathop{\lim }}\,\frac{\sqrt{f(x)}-1}{\sqrt{x}-1}\] Þ \[y=\underset{x\to 1}{\mathop{\lim }}\,\frac{\left( \sqrt{f(x)}-1 \right)}{\left( \sqrt{x}-1 \right)}.\frac{\left( \sqrt{f(x)}+1 \right)}{\left( \sqrt{x}+1 \right)}.\frac{\left( \sqrt{x}+1 \right)}{\left( \sqrt{f(x)}+1 \right)}\] Þ \[y=\underset{x\to 1}{\mathop{\lim }}\,\frac{f(x)-1}{x-1}.\frac{\sqrt{x}+1}{\sqrt{f(x)}+1}\] Þ \[y=\underset{x\to 1}{\mathop{\lim }}\,\frac{f(x)-f(1)}{x-1}.\underset{x\to 1}{\mathop{\lim }}\,\frac{\sqrt{x}+1}{\sqrt{f(x)}+1}\] Þ \[y={f}'(1).\frac{2}{\sqrt{f(1)}+1}\]Þ \[y=2.\frac{2}{2}=2\] Trick : Applying L-Hospital?s rule, \[\underset{x\to 1}{\mathop{\lim }}\,\frac{\frac{1}{2}{{\left\{ f(x) \right\}}^{-1/2}}{f}'(x)}{\frac{1}{2}{{x}^{-1/2}}}=\underset{x\to 1}{\mathop{\lim }}\,\frac{{f}'(x)\sqrt{x}}{\sqrt{f(x)}}=\frac{{f}'(1).\sqrt{1}}{\sqrt{f(1)}}=2.\]
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