mathop {lim }limits_{x to - 1} frac{{sqrt pi | vedclass

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

Question

(B) Let ${\cos ^{ - 1}}x = y$. As $x 	o - 1$,$y 	o \pi$. Then,the limit becomes $\mathop {\lim }\limits_{y 	o \pi } \frac{{\sqrt \pi - \sqrt y }}{{

Vedclass · Accepted Answer

$\frac{1}{{\sqrt {2\pi } }}$

Vedclass · Answer

(B) Let ${\cos ^{ - 1}}x = y$. As $x 	o - 1$,$y 	o \pi$. Then,the limit becomes $\mathop {\lim }\limits_{y 	o \pi } \frac{{\sqrt \pi - \sqrt y }}{{\sqrt {1 + \cos y} }}$. Using the identity $1 + \cos y = 2 \cos^2(y/2)$,we have $\sqrt{1 + \cos y} = \sqrt{2} |\cos(y/2)|$. Since $y 	o \pi$,$\cos(y/2) > 0$,so $\sqrt{1 + \cos y} = \sqrt{2} \cos(y/2)$. Now,$\cos(y/2) = \sin(\pi/2 - y/2)$. So the limit is $\mathop {\lim }\limits_{y 	o \pi } \frac{{\sqrt \pi - \sqrt y }}{{\sqrt{2} \sin(\frac{\pi - y}{2})}}$. Multiply numerator and denominator by $(\sqrt{\pi} + \sqrt{y})$ and use the standard limit $\mathop {\lim }\limits_{	heta 	o 0} \frac{\sin 	heta}{	heta} = 1$: $= \mathop {\lim }\limits_{y 	o \pi } \frac{\pi - y}{\sqrt{2} \sin(\frac{\pi - y}{2}) (\sqrt{\pi} + \sqrt{y})} = \mathop {\lim }\limits_{y 	o \pi } \frac{2 \cdot \frac{\pi - y}{2}}{\sqrt{2} \sin(\frac{\pi - y}{2}) (\sqrt{\pi} + \sqrt{y})} = \frac{2}{\sqrt{2} (2\sqrt{\pi})} = \frac{1}{\sqrt{2\pi}}$.

$\mathop {\lim }\limits_{x \to - 1} \frac{{\sqrt \pi - \sqrt {{{\cos }^{ - 1}}x} }}{{\sqrt {x + 1} }}$ is equal to

Similar Questions

If $\lim _{n \rightarrow \infty} x_n$ exists and is finite,$x_1=2$,$x_{n+1}=\frac{a+b x_n}{b+c x_n}$ for all $n \in N$,and $c > b > a > 0$,then $\lim _{n \rightarrow \infty} x_n =$

If $[t]$ represents the greatest integer $\leq t$,then the value of $\lim _{x \rightarrow 3} \frac{11-[2-x]}{[x+10]}$ is

$\mathop {\lim }\limits_{x \to a} \frac{{{x^2} - {a^2}}}{{x - a}} = $

If $\lim_{x \to 0} \frac{45^x - 9^x - 5^x + 1}{(kx - 1)(3^x - 1)} = 2$, then the value of $k$ is...

The value of $\mathop {Limit}\limits_{x \to \infty } \,\frac{{{{\left( {{2^{{x^n}}}} \right)}^{\frac{1}{{{e^x}}}}}\,\, - \,\,{{\left( {{3^{{x^n}}}} \right)}^{\frac{1}{{{e^x}}}}}}}{{{x^n}}}\,$ (where $n \in N$) is

Vedclass Test Series

Exam Paper Generator

Online Exam Module