If f(1) = 1 and f'(1) = 4, then the value of | vedclass

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

Question

(B) Given the limit $L = \mathop {\lim }\limits_{x 	o 1} \frac{{\sqrt {f(x)} - 1}}{{\sqrt x - 1}}$.Since the form is $\frac{0}{0}$,we apply $L$'$H$ôp

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Given the limit $L = \mathop {\lim }\limits_{x 	o 1} \frac{{\sqrt {f(x)} - 1}}{{\sqrt x - 1}}$.Since the form is $\frac{0}{0}$,we apply $L$'$H$ôpital's rule:$L = \mathop {\lim }\limits_{x 	o 1} \frac{\frac{d}{dx}(\sqrt{f(x)} - 1)}{\frac{d}{dx}(\sqrt{x} - 1)}$$L = \mathop {\lim }\limits_{x 	o 1} \frac{\frac{f'(x)}{2\sqrt{f(x)}}}{\frac{1}{2\sqrt{x}}}$$L = \mathop {\lim }\limits_{x 	o 1} \frac{f'(x) \cdot \sqrt{x}}{\sqrt{f(x)}}$Substituting $x = 1$:$L = \frac{f'(1) \cdot \sqrt{1}}{\sqrt{f(1)}} = \frac{4 \cdot 1}{1} = 4$.

If $f(1) = 1$ and $f'(1) = 4,$ then the value of $\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {f(x)} - 1}}{{\sqrt x - 1}}$ is

Similar Questions

$\mathop {\lim }\limits_{x \to 0} \frac{{{2^x} - 1}}{{{{(1 + x)}^{1/2}} - 1}} = $

The value of $\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{ - x}}}}{{\sin x}}$ is

$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}{x} = $

$\mathop {\lim }\limits_{x \to 0} \frac{{x\cos x - \sin x}}{{{x^2}\sin x}} = $

Let $f: R \rightarrow R$ be a function such that $f(2)=4$ and $f^{\prime}(2)=1$. Then,the value of $\lim _{x \rightarrow 2} \frac{x^{2} f(2)-4 f(x)}{x-2}$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module