mathop {lim }limits_{x to infty } frac{{{{cot | vedclass

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

Question

(A) माना $L = \lim _{x}$ ${\rightarrow \infty} \frac{\cot ^{-1}(\sqrt{x+1}-\sqrt{x})}{\sec ^{-1}\left\{\left(\frac{2 x+1}{x-1}\right)^{x}\right\}}$.पह

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) माना $L = \lim _{x}$ ${ightarrow \infty} \frac{\cot ^{-1}(\sqrt{x+1}-\sqrt{x})}{\sec ^{-1}\left\{\left(\frac{2 x+1}{x-1}ight)^{x}ight\}}$.पहले,अंश को सरल करने पर:$\sqrt{x+1}-\sqrt{x} = \frac{1}{\sqrt{x+1}+\sqrt{x}}$.जैसे $x 	o \infty$,$\frac{1}{\sqrt{x+1}+\sqrt{x}} 	o 0$,इसलिए $\cot ^{-1}(0) = \frac{\pi}{2}$.अब,हर की जाँच करने पर:जैसे $x 	o \infty$,$\frac{2x+1}{x-1} 	o 2$. अतः $(\frac{2x+1}{x-1})^x 	o \infty$.और $\sec ^{-1}(\infty) = \frac{\pi}{2}$.अतः,$L = \frac{\pi/2}{\pi/2} = 1$.

$\mathop {\lim }\limits_{x \to \infty } \frac{{{{\cot }^{ - 1}}\left( {\sqrt {x + 1} - \sqrt x } \right)}}{{{{\sec }^{ - 1}}\left\{ {{{\left( {\frac{{2x + 1}}{{x - 1}}} \right)}^x}} \right\}}}$ का मान ज्ञात कीजिए।

Similar Questions

$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{1/x}} - 1}}{{{e^{1/x}} + 1}} = $

$\operatorname{Lt}_{x \rightarrow 0} \left( \frac{1+5x^2}{1+3x^2} \right)^{\frac{1}{x^2}}$ का मान है

$\lim _{x \rightarrow \infty}\left(1+\frac{2}{x}\right)^{3 x}=$

यदि $x > 2$ के लिए $g(x) = \frac{x}{[x]}$ है,तो $\lim_{x \rightarrow 2^+} \frac{g(x) - g(2)}{x - 2}$ का मान ज्ञात कीजिए।

$\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\left[ {1 - \tan \left( {\frac{x}{2}} \right)} \right]\,[1 - \sin x]}}{{\left[ {1 + \tan \left( {\frac{x}{2}} \right)} \right]\,{{[\pi - 2x]}^3}}}$ का मान है

Vedclass Test Series

Exam Paper Generator

Online Exam Module