lim _{x rightarrow frac{pi}{4}} frac{8 sqrt{2 | vedclass

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

Question

(A) माना $L = \lim _{x \rightarrow \frac{\pi}{4}} \frac{8 \sqrt{2}-(\cos x+\sin x)^{7}}{\sqrt{2}-\sqrt{2} \sin 2 x}$। यह $\frac{0}{0}$ रूप है।$L'H\hat

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(A) माना $L = \lim _{x ightarrow \frac{\pi}{4}} \frac{8 \sqrt{2}-(\cos x+\sin x)^{7}}{\sqrt{2}-\sqrt{2} \sin 2 x}$। यह $\frac{0}{0}$ रूप है।$L'H\hat{o}pital$ नियम लागू करने पर:$L = \lim _{x ightarrow \frac{\pi}{4}} \frac{-7(\cos x+\sin x)^{6}(-\sin x+\cos x)}{-2 \sqrt{2} \cos 2 x}$ध्यान दें कि $\cos 2x = \cos^2 x - \sin^2 x = (\cos x - \sin x)(\cos x + \sin x)$।यह मान रखने पर:$L = \lim _{x}$ ${ightarrow \frac{\pi}{4}} \frac{-7(\cos x+\sin x)^{6}(\cos x - \sin x)}{-2 \sqrt{2} (\cos x - \sin x)(\cos x + \sin x)}$$L = \lim _{x ightarrow \frac{\pi}{4}} \frac{7(\cos x+\sin x)^{5}}{2 \sqrt{2}}$$x = \frac{\pi}{4}$ पर,$(\cos x + \sin x) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$।$L = \frac{7(\sqrt{2})^{5}}{2 \sqrt{2}} = \frac{7 	imes 4 \sqrt{2}}{2 \sqrt{2}} = \frac{28}{2} = 14$।

$\lim _{x \rightarrow \frac{\pi}{4}} \frac{8 \sqrt{2}-(\cos x+\sin x)^{7}}{\sqrt{2}-\sqrt{2} \sin 2 x}$ का मान ज्ञात कीजिए।

Similar Questions

$\lim _{x \rightarrow 1} \frac{\log x}{1-x} = $

मान लीजिए $f: R \rightarrow R$, $x=0$ पर अवकलनीय है। यदि $f(0)=0$ और $f'(0)=2$ है, तो $\lim _{x \rightarrow 0} \frac{1}{x} [f(x)+f(2 x)+f(3 x)+\ldots+f(2015 x)]$ का मान ज्ञात कीजिए।

$\mathop {\lim }\limits_{n \to \infty } {\left[ {\frac{{\sin \left( n \right)}}{{{n^2}}} + \log \left( {\frac{{en + 1}}{{n + e}}} \right)} \right]^n}$ का मान ज्ञात कीजिए।

$\mathop {\lim }\limits_{\alpha \to \beta } \left[ {\frac{{{{\sin }^2}\alpha - {{\sin }^2}\beta }}{{{\alpha ^2} - {\beta ^2}}}} \right] = $

$\mathop {\lim }\limits_{x \to 0} \left[ {\frac{1}{x} - \frac{{\log (1 + x)}}{{{x^2}}}} \right] =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module