mathop {lim }limits_{x to - 2} frac{{{{sin }^ | vedclass

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

Question

(C) Let $y = \mathop {\lim }\limits_{x 	o - 2} \frac{{{{\sin }^{ - 1}}(x + 2)}}{{{x^2} + 2x}}$.Since the limit is in the $\frac{0}{0}$ form,we apply

Vedclass · Accepted Answer

$-1/2$

Vedclass · Answer

(C) Let $y = \mathop {\lim }\limits_{x 	o - 2} \frac{{{{\sin }^{ - 1}}(x + 2)}}{{{x^2} + 2x}}$.Since the limit is in the $\frac{0}{0}$ form,we apply $L'	ext{Hospital's rule}$.$y = \mathop {\lim }\limits_{x 	o - 2} \frac{\frac{d}{dx}(\sin^{-1}(x+2))}{\frac{d}{dx}(x^2+2x)}$$y = \mathop {\lim }\limits_{x 	o - 2} \frac{\frac{1}{\sqrt{1-(x+2)^2}}}{2x+2}$Substituting $x = -2$:$y = \frac{\frac{1}{\sqrt{1-0}}}{2(-2)+2} = \frac{1}{-4+2} = -\frac{1}{2}$.

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

Similar Questions

The value of $\mathop {\lim }\limits_{x \to 7} \frac{{2 - \sqrt {x - 3} }}{{{x^2} - 49}}$ is

If $\alpha = \lim_{x \rightarrow \pi/4} \frac{\tan^{3} x - \tan x}{\cos(x + \pi/4)}$ and $\beta = \lim_{x \rightarrow 0} (\cos x)^{\cot x}$ are the roots of the equation $ax^{2} + bx - 4 = 0$,then the ordered pair $(a, b)$ is:

Evaluate $\lim _{x \rightarrow 0^{+}} (x^{n} \ln x)$ for $n > 0$.

Evaluate the limit: $\lim _{x \rightarrow 0} \frac{e^x-e^{\sin x}}{2(x-\sin x)}$

$\lim _{x \rightarrow 0} \frac{x^2 2^x-x^2 \sin x-x^2}{3^x+\cos x-3^x \cos x-1}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module