If f (1) = 1, f1(1) = 2, then displaystyle limx → 1 (√f(x) - 1/ √x -1) is

Question

If f (1) = 1, f1(1) = 2, then displaystyle limx → 1 (√f(x) - 1/ √x -1) is (A) 2 (B) 4 (C) 1 (D) 1/2

Tardigrade Admin · Accepted Answer

displaystyle limx → 1 (√f(x) -1/√x -1) ((0/0)) form using Lâ Hospitalâs rule = displaystyle lim x → 1 ((1/2√f(x))f'(x)/1 /2 √x) = (f'(1)/√f(1)) = (2/1) = 2 .

Q. If $f (1) = 1$ , $f^{1} (1) = 2$ , then $x \to 1 lim \frac{f ( x ) - 1}{x - 1}$ is

2

4

1

1/2

$x \to 1 lim \frac{f ( x ) - 1}{x - 1} (\frac{0}{0})$ form using L’ Hospital’s rule
$= x \to 1 lim \frac{\frac{1}{2 f ( x )} f ^{'} ( x )}{1/2 x} = \frac{f ^{'} ( 1 )}{f ( 1 )} = \frac{2}{1} = 2$ .

Q. If f(1)=1, f1(1)=2, then x→1lim​x​−1f(x)​−1​ is

2

4

1

1/2

x→1lim​x​−1f(x)​−1​(00​) form using L’ Hospital’s rule =x→1lim​1/2x​2f(x)​1​f′(x)​=f(1)​f′(1)​=12​=2.

Q. If $f (1) = 1$ , $f^{1} (1) = 2$ , then $x \to 1 lim \frac{f ( x ) - 1}{x - 1}$ is

$x \to 1 lim \frac{f ( x ) - 1}{x - 1} (\frac{0}{0})$ form using L’ Hospital’s rule
$= x \to 1 lim \frac{\frac{1}{2 f ( x )} f ^{'} ( x )}{1/2 x} = \frac{f ^{'} ( 1 )}{f ( 1 )} = \frac{2}{1} = 2$ .