mathop {Limit}limits_{x to 0^+} frac{1}{xsqrt | vedclass

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

Question

(D) Let $f(x) = a 	an^{-1} \frac{\sqrt{x}}{a} - b 	an^{-1} \frac{\sqrt{x}}{b}$ and $g(x) = x\sqrt{x} = x^{3/2}$.Since the limit is of the form $0/0$

Vedclass · Accepted Answer

$\frac{a^2 - b^2}{3a^2b^2}$

Vedclass · Answer

(D) Let $f(x) = a 	an^{-1} \frac{\sqrt{x}}{a} - b 	an^{-1} \frac{\sqrt{x}}{b}$ and $g(x) = x\sqrt{x} = x^{3/2}$.Since the limit is of the form $0/0$,we apply $L$'Hopital's rule:$\frac{d}{dx} (a 	an^{-1} \frac{\sqrt{x}}{a}) = a \cdot \frac{1}{1 + (\sqrt{x}/a)^2} \cdot \frac{1}{2a\sqrt{x}} = \frac{a^2}{a^2 + x} \cdot \frac{1}{2\sqrt{x}}$.Similarly,$\frac{d}{dx} (b 	an^{-1} \frac{\sqrt{x}}{b}) = \frac{b^2}{b^2 + x} \cdot \frac{1}{2\sqrt{x}}$.Derivative of $g(x) = \frac{3}{2}\sqrt{x}$.Applying the rule: $\lim_{x 	o 0^+} \frac{\frac{1}{2\sqrt{x}} (\frac{a^2}{a^2+x} - \frac{b^2}{b^2+x})}{\frac{3}{2}\sqrt{x}} = \lim_{x 	o 0^+} \frac{1}{3x} \left( \frac{a^2(b^2+x) - b^2(a^2+x)}{(a^2+x)(b^2+x)} ight)$.$= \lim_{x 	o 0^+} \frac{1}{3x} \left( \frac{a^2b^2 + a^2x - b^2a^2 - b^2x}{(a^2+x)(b^2+x)} ight) = \lim_{x 	o 0^+} \frac{1}{3x} \left( \frac{x(a^2 - b^2)}{(a^2+x)(b^2+x)} ight)$.$= \frac{a^2 - b^2}{3a^2b^2}$.

$\mathop {Limit}\limits_{x \to 0^+} \frac{1}{x\sqrt{x}} \left( a \tan^{-1} \frac{\sqrt{x}}{a} - b \tan^{-1} \frac{\sqrt{x}}{b} \right)$ has the value equal to

Similar Questions

$\mathop {\lim }\limits_{x \to \frac{\pi }{2}} (1 - \sin x)\tan x$ is

$\mathop {\lim }\limits_{x \to \pi /2} \left[ {x\tan x - \left( {\frac{\pi }{2}} \right)\sec x} \right] = $

If $f(3) = 6$ and $f'(3) = 2$,then $\mathop {\text{Limit}}\limits_{x \to 3} \frac{x f(3) - 3 f(x)}{x - 3}$ is given by:

$\mathop {\lim }\limits_{x \to 0} \frac{{\sin x + \log (1 - x)}}{{{x^2}}}$ is equal to

$\mathop {\lim }\limits_{x \to 0} \frac{{\tan 2x - x}}{{3x - \sin x}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module