lim _{x} {rightarrow frac{pi}{2}} frac{left(1 | vedclass

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

Question

(B) ધારો કે $l = \lim _{x}$ ${ightarrow \frac{\pi}{2}} \left[ \frac{1-	an \left(\frac{x}{2}ight)}{1+	an \left(\frac{x}{2}ight)} ight] \left[

Vedclass · Accepted Answer

$\frac{1}{32}$

Vedclass · Answer

(B) ધારો કે $l = \lim _{x}$ ${ightarrow \frac{\pi}{2}} \left[ \frac{1-	an \left(\frac{x}{2}ight)}{1+	an \left(\frac{x}{2}ight)} ight] \left[ \frac{1-\sin x}{(\pi-2 x)^3} ight]$ $	an(\frac{\pi}{4} - \frac{x}{2}) = \frac{1-	an(x/2)}{1+	an(x/2)}$ હોવાથી: $l = \lim _{x ightarrow \frac{\pi}{2}} \frac{	an \left(\frac{\pi}{4}-\frac{x}{2}ight)(1-\sin x)}{(\pi-2 x)^3}$ $\pi-2x = 	heta$ લેતા,$x = \frac{\pi}{2} - \frac{	heta}{2}$. જ્યારે $x ightarrow \frac{\pi}{2}$,ત્યારે $	heta ightarrow 0$. વળી,$\frac{\pi}{4} - \frac{x}{2} = \frac{	heta}{4}$ અને $1-\sin x = 2\sin^2(\frac{	heta}{4})$. કિંમતો મૂકતા: $l = \lim _{	heta ightarrow 0} \frac{	an(\frac{	heta}{4}) \cdot 2\sin^2(\frac{	heta}{4})}{	heta^3}$ $l = \lim _{	heta}$ ${ightarrow 0} \frac{	an(\frac{	heta}{4})}{\frac{	heta}{4} \cdot 4} \cdot \frac{2\sin^2(\frac{	heta}{4})}{(\frac{	heta}{4})^2 \cdot 16}$ $l = \frac{2}{64} (1)(1)^2 = \frac{1}{32}$

$\lim _{x}$ ${\rightarrow \frac{\pi}{2}} \frac{\left(1-\tan \left(\frac{x}{2}\right)\right)(1-\sin x)}{\left(1+\tan \left(\frac{x}{2}\right)\right)(\pi-2 x)^3}$ ની કિંમત શોધો.

Similar Questions

જો $\mathop {\lim }\limits_{x \to a} \frac{{{x^9} + {a^9}}}{{x + a}} = 9$ હોય,તો $a = $

$\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{2}{x}} \right)^x} = $

$\lim _{x \rightarrow 0} \frac{\sin ^2 x}{\sqrt{2}-\sqrt{1+\cos x}}$ ની કિંમત શોધો.

$\lim _{x}$ ${\rightarrow 1} \frac{(1-x)(1-x^2) \cdots (1-x^{2n})}{\{(1-x)(1-x^2) \cdots (1-x^n)\}^2} = \dots, \forall n \in N$

$\mathop {\lim }\limits_{x \to 0} \frac{{\tan x - \sin x}}{{{x^3}}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module