mathop {Limit}limits_{x to infty } ,frac{{{{c | vedclass

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

Question

(A) First,evaluate the numerator: $\mathop {Limit}\limits_{x 	o \infty } (\sqrt {x + 1} - \sqrt x) = \mathop {Limit}\limits_{x 	o \infty } \frac{(x+

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) First,evaluate the numerator: $\mathop {Limit}\limits_{x 	o \infty } (\sqrt {x + 1} - \sqrt x) = \mathop {Limit}\limits_{x 	o \infty } \frac{(x+1)-x}{\sqrt{x+1}+\sqrt{x}} = \mathop {Limit}\limits_{x 	o \infty } \frac{1}{\sqrt{x+1}+\sqrt{x}} = 0$. Thus,$\cot^{-1}(0) = \pi / 2$. Next,evaluate the denominator: $\mathop {Limit}\limits_{x 	o \infty } (\frac{2x+1}{x-1})^x = \mathop {Limit}\limits_{x 	o \infty } (2 + \frac{3}{x-1})^x = \infty$. Thus,$\sec^{-1}(\infty) = \pi / 2$. Therefore,the limit is $\frac{\pi / 2}{\pi / 2} = 1$.

$\mathop {Limit}\limits_{x \to \infty } \,\frac{{{{\cot }^{ - 1}}\left( {\sqrt {x + 1} \, - \,\sqrt x } \right)}}{{{{\sec }^{ - 1}}\left\{ {{{\left( {\frac{{2x + 1}}{{x - 1}}} \right)}^x}} \right\}}}$ is equal to

Similar Questions

The imaginary part of $\tan^{-1}\left(\frac{5i}{3}\right)$ is

If the equation $2 \operatorname{Cot}^{-1}(x^2+2x+k) = \pi - 3 \operatorname{Tan}^{-1}(x^2+2x+k)$ has two distinct real solutions,then all the values of $k$ lie in the interval

The number of solutions of $\sin ^{-1} x+\sin ^{-1}(1-x)=\cos ^{-1} x$ is

If $y=\cos ^{-1}\left(\frac{a^2-x^2}{a^2+x^2}\right)+\sin ^{-1}\left(\frac{2 a x}{a^2+x^2}\right)$,then $\frac{d y}{d x}$ is equal to

Find $\frac{dy}{dx}$ if $y = \sin^{-1}\left(\frac{1-x^2}{1+x^2}\right)$,where $0 < x < 1$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module