Question

30. Iff is differentiable at x = 1, then lim x_1 (x²f(1)-f(x))/x-1
is
(A) -f'(1)
(B) f(1) - f'(1) (C) 2f (1) - f'(1)
(D) 2f (1) + f'(1) (E) f(1) + f'(1)​

Answer 1

Given:

f(x) is a differentiable at x = 1

To Find :

The value of

$\bf\blue{\lim_{x\to\:1}\dfrac{x^2f(1)-f(x)}{x-1}}$

Solution :

We have to find the value of

$\sf\lim_{x\to\:1}\dfrac{x^2f(1)-f(x)}{x-1}$

Clearly it is a 0/0 form limit

By L'Hospital rule, then

$\sf\lim_{x\to\:1}\dfrac{\frac{(x^2f(1)-f(x))}{dx}}{\frac{d(x-1)}{dx}}$

Then ,

$\sf\lim_{x\to\:1}\dfrac{2xf(1)-f'(x)}{1}$

$\sf=2f(1)-f'(1)$

Hence, C option is correct!