Functions Question 260
Question: If $ f(1),=1,,{f}’,(1),=2 $ , then $ \underset{x\to 1}{\mathop{\lim }},\frac{\sqrt{f(x)}-1}{\sqrt{x}-1} $ is
[AIEEE 2002]
Options:
A) 2
B) 4
C) 1
D) ½
Show Answer
Answer:
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{( \sqrt{f(x)}-1 )}{( \sqrt{x}-1 )}.\frac{( \sqrt{f(x)}+1 )}{( \sqrt{x}+1 )}.\frac{( \sqrt{x}+1 )}{( \sqrt{f(x)}+1 )} $
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}{{{ f(x) }}^{-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. $
BETA
