If f(2) = 4 and f'(2) = 1,then mathop {lim }l | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $f(2) = 4$ and $f'(2) = 1$.We need to evaluate $\mathop {\lim }\limits_{x 	o 2} \frac{xf(2) - 2f(x)}{x - 2}$.Adding and subtracting $2f(2)$

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Given $f(2) = 4$ and $f'(2) = 1$.We need to evaluate $\mathop {\lim }\limits_{x 	o 2} \frac{xf(2) - 2f(x)}{x - 2}$.Adding and subtracting $2f(2)$ in the numerator:$= \mathop {\lim }\limits_{x 	o 2} \frac{xf(2) - 2f(2) + 2f(2) - 2f(x)}{x - 2}$$= \mathop {\lim }\limits_{x 	o 2} \left[ \frac{f(2)(x - 2)}{x - 2} - \frac{2(f(x) - f(2))}{x - 2} ight]$$= f(2) - 2 \mathop {\lim }\limits_{x 	o 2} \frac{f(x) - f(2)}{x - 2}$$= f(2) - 2f'(2)$Substituting the values: $4 - 2(1) = 4 - 2 = 2$.Alternatively,using $L$-Hospital rule:$= \mathop {\lim }\limits_{x 	o 2} \frac{\frac{d}{dx}(xf(2) - 2f(x))}{\frac{d}{dx}(x - 2)}$$= \mathop {\lim }\limits_{x 	o 2} \frac{f(2) - 2f'(x)}{1} = f(2) - 2f'(2) = 4 - 2(1) = 2$.

If $f(2) = 4$ and $f'(2) = 1$,then $\mathop {\lim }\limits_{x \to 2} \frac{xf(2) - 2f(x)}{x - 2} = $

Similar Questions

For constant $a$,$\frac{d}{d x}\left(x^{x}+x^{a}+a^{x}+a^{a}\right)$ is

If $f(x) = \frac{1+\sec x}{2(\sec x-1)}$ for $0 < x < \frac{\pi}{2}$ and $f^{\prime}(x) = f(x) \cdot g(x)$, then $g(x) =$

Differentiate the function with respect to $x$: $\cos(x^{3}) \cdot \sin^{2}(x^{5})$

Find the derivative of the following function (it is to be understood that $a, b, c, d, p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $\frac{a+b \sin x}{c+d \cos x}$

Derivative of $e^x$ with respect to $\sqrt{x}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module