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Limits Continuity and Differentiability question

2002 · Shift 0 · Q69

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JEE Main
Mathematics
Limits Continuity and Differentiability
2002 · Shift 0 · Q69

Limits Continuity and Differentiability question

2002 · Shift 0 · Q69

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1

If

f\left( 1 \right) = 1,{f'}\left( 1 \right) = 2,

then

\mathop {\lim }\limits_{x \to 1} {{\sqrt {f\left( x \right)} - 1} \over {\sqrt x - 1}}

is

A
$2$
B
$4$
C
$1$
D
${1 \over 2}$

View written solutionFree

Correct answer: A

We need to evaluate

L=\lim_{x\to 1}\frac{\sqrt{f(x)}-1}{\sqrt{x}-1}

given that

f(1)=1,\qquad f'(1)=2.

Since $f(1)=1$ , as $x\to 1$ we have

\sqrt{f(x)}-1 \to 0,

and also

\sqrt{x}-1\to 0.

So this is a $\frac{0}{0}$ form.

Rationalize both numerator and denominator:

\sqrt{f(x)}-1=\frac{f(x)-1}{\sqrt{f(x)}+1},

\sqrt{x}-1=\frac{x-1}{\sqrt{x}+1}.

Hence,

L=\lim_{x\to 1}\frac{\dfrac{f(x)-1}{\sqrt{f(x)}+1}}{\dfrac{x-1}{\sqrt{x}+1}}.

Simplify:

L=\lim_{x\to 1}\frac{f(x)-1}{x-1}\cdot \frac{\sqrt{x}+1}{\sqrt{f(x)}+1}.

Now use the given information. As $x\to 1$ ,

\frac{f(x)-1}{x-1}\to f'(1)=2.

Also,

\sqrt{x}+1\to 2,

and since $f(x)\to f(1)=1$ ,

\sqrt{f(x)}+1\to 2.

Therefore,

L=2\cdot \frac{2}{2}=2.

So the correct option is

\boxed{2}

which is option A.

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