If f(0) = 2, then mathop {lim }limits_{x to 0 | vedclass

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

Question

(C) Let $L = \mathop {\lim }\limits_{x 	o 0} \frac{{\int\limits_0^x {tf(x)dt + \int\limits_0^x xf(t)dt} }}{{{x^2}}}$$L = \mathop {\lim }\limits_{x

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) Let $L = \mathop {\lim }\limits_{x 	o 0} \frac{{\int\limits_0^x {tf(x)dt + \int\limits_0^x xf(t)dt} }}{{{x^2}}}$$L = \mathop {\lim }\limits_{x 	o 0} \frac{{f(x) \int\limits_0^x t dt + x \int\limits_0^x f(t) dt}}{{{x^2}}}$$L = \mathop {\lim }\limits_{x 	o 0} \frac{{f(x) \cdot \frac{x^2}{2} + x \int\limits_0^x f(t) dt}}{{{x^2}}}$$L = \mathop {\lim }\limits_{x 	o 0} \left( \frac{f(x)}{2} + \frac{\int\limits_0^x f(t) dt}{x} ight)$Using $L$'Hopital's rule for the second term: $\mathop {\lim }\limits_{x 	o 0} \frac{\int\limits_0^x f(t) dt}{x} = \mathop {\lim }\limits_{x 	o 0} \frac{f(x)}{1} = f(0)$$L = \frac{f(0)}{2} + f(0) = \frac{3}{2} f(0)$Given $f(0) = 2$,$L = \frac{3}{2} 	imes 2 = 3$.

If $f(0) = 2,$ then $\mathop {\lim }\limits_{x \to 0} \frac{{\int\limits_0^x {\left( {tf(x) + xf(t)} \right)dt} }}{{{x^2}}}$ is equal to -

Similar Questions

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

$\mathop {\lim }\limits_{x \to 1} \frac{{\log x}}{{x - 1}} = $

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

Let $\alpha$ and $\beta$ be real numbers such that $\lim _{x \rightarrow 0} \frac{1}{x^3}\left(\frac{\alpha}{2} \int_0^x \frac{1}{1-t^2} d t+\beta x \cos x\right)=2$. Then the value of $\alpha+\beta$ is $....$ (in $.40$)

$\lim\limits_{x \rightarrow 0} \frac{\int\limits_{0}^{x} t \sin (10 t) d t}{x}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module