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

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) ધારો કે $L = \lim _{x}$ ${ightarrow \infty} \frac{\cot ^{-1}(\sqrt{x+1}-\sqrt{x})}{\sec ^{-1}\left\{\left(\frac{2 x+1}{x-1}ight)^{x}ight\}}$.પ્રથમ,અંશનું સાદું રૂપ આપતા:$\sqrt{x+1}-\sqrt{x} = \frac{1}{\sqrt{x+1}+\sqrt{x}}$.જેમ $x 	o \infty$,તેમ $\frac{1}{\sqrt{x+1}+\sqrt{x}} 	o 0$,તેથી $\cot ^{-1}(0) = \frac{\pi}{2}$.હવે,છેદ તપાસતા:જેમ $x 	o \infty$,તેમ $\frac{2x+1}{x-1} 	o 2$. તેથી $(\frac{2x+1}{x-1})^x 	o \infty$.અને $\sec ^{-1}(\infty) = \frac{\pi}{2}$.તેથી,$L = \frac{\pi/2}{\pi/2} = 1$.

$\mathop {\lim }\limits_{x \to \infty } \frac{{{{\cot }^{ - 1}}\left( {\sqrt {x + 1} - \sqrt x } \right)}}{{{{\sec }^{ - 1}}\left\{ {{{\left( {\frac{{2x + 1}}{{x - 1}}} \right)}^x}} \right\}}}$ ની કિંમત શોધો.

Similar Questions

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

ધારો કે $f(x) = \frac{\sqrt{x+3}}{x+1}$ છે, તો $\lim_{x \rightarrow -3^{-}} f(x)$ ની કિંમત શોધો.

ધારો કે $f(x) = \frac{x \cdot 2^x - x}{1 - \cos x}$ અને $g(x) = 2^x \sin \left( \frac{\ln 2}{2^x} \right)$,તો:

$\lim \limits_{x}$ ${\rightarrow \frac{1}{\sqrt{2}}} \frac{\sin \left(\cos ^{-1} x\right)-x}{1-\tan \left(\cos ^{-1} x\right)}$ ની કિંમત શોધો.

$\mathop {\lim }\limits_{x \to 1} f(x)$ ની કિંમત શોધો,જ્યાં $f(x) = \begin{cases} \frac{e^{\frac{1}{x-1}} - 2}{e^{\frac{1}{x-1}} + 2} & x \neq 1 \\ 1 & x = 1 \end{cases}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module