lim limits_{x} {rightarrow frac{1}{sqrt{2}}} | vedclass

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

Question

(D) माना $f(x) = \frac{\sin(\cos^{-1} x) - x}{1 - 	an(\cos^{-1} x)}$.हम जानते हैं कि $\cos^{-1} x = \sin^{-1}(\sqrt{1-x^2})$ और $\cos^{-1} x = 	an^{

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) माना $f(x) = \frac{\sin(\cos^{-1} x) - x}{1 - 	an(\cos^{-1} x)}$.हम जानते हैं कि $\cos^{-1} x = \sin^{-1}(\sqrt{1-x^2})$ और $\cos^{-1} x = 	an^{-1}(\frac{\sqrt{1-x^2}}{x})$.इन मानों को व्यंजक में रखने पर:$\lim \limits_{x}$ ${ightarrow \frac{1}{\sqrt{2}}} \frac{\sin(\sin^{-1}(\sqrt{1-x^2})) - x}{1 - 	an(	an^{-1}(\frac{\sqrt{1-x^2}}{x}))}$$= \lim \limits_{x ightarrow \frac{1}{\sqrt{2}}} \frac{\sqrt{1-x^2} - x}{1 - \frac{\sqrt{1-x^2}}{x}}$$= \lim \limits_{x ightarrow \frac{1}{\sqrt{2}}} \frac{\sqrt{1-x^2} - x}{\frac{x - \sqrt{1-x^2}}{x}}$$= \lim \limits_{x ightarrow \frac{1}{\sqrt{2}}} \frac{- (x - \sqrt{1-x^2})}{\frac{x - \sqrt{1-x^2}}{x}}$$= \lim \limits_{x ightarrow \frac{1}{\sqrt{2}}} (-x) = -\frac{1}{\sqrt{2}}$.

$\lim \limits_{x}$ ${\rightarrow \frac{1}{\sqrt{2}}} \frac{\sin \left(\cos ^{-1} x\right)-x}{1-\tan \left(\cos ^{-1} x\right)}$ का मान ज्ञात कीजिए।

Similar Questions

$\lim _{x \rightarrow a} \frac{\sqrt{a+2 x}-\sqrt{3 x}}{\sqrt{3 a+x}-2 \sqrt{x}}$ का मान है

$\lim _{n}$ ${\rightarrow \infty} n^{-n k} \left\{(n+1)\left(n+\frac{1}{2}\right)\left(n+\frac{1}{2^2}\right) \ldots\left(n+\frac{1}{2^{k-1}}\right)\right\}^n=$

यदि $a = \lim_{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{n^2}$ और $b = \lim_{n \rightarrow \infty} \frac{1^2+2^2+3^2+\ldots+n^2}{n^3}$ है,तो

दिए गए सीमा (limit) का मूल्यांकन करें: $\mathop {\lim }\limits_{x \to -1} \frac{x^{10}+x^{5}+1}{x-1}$

$\lim _{x \rightarrow 0} \frac{1-\cos \left(x^2+\pi(x+2)\right)}{x^2} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module