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

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

Question

(A) Let $L = \lim _{x \rightarrow \frac{\pi}{4}} \frac{4 \sqrt{2}-(\cos x+\sin x)^5}{1-\sin 2 x}$. This is a $\frac{0}{0}$ form.Applying $L'H\hat{o}pi

Vedclass · Accepted Answer

$5 \sqrt{2}$

Vedclass · Answer

(A) Let $L = \lim _{x \rightarrow \frac{\pi}{4}} \frac{4 \sqrt{2}-(\cos x+\sin x)^5}{1-\sin 2 x}$. This is a $\frac{0}{0}$ form.Applying $L'H\hat{o}pital's$ Rule:$L = \lim _{x \rightarrow \frac{\pi}{4}} \frac{-5(\cos x+\sin x)^4(-\sin x+\cos x)}{-2 \cos 2 x}$.Using the identity $1-\sin 2x = (\cos x - \sin x)^2$ and $\cos 2x = (\cos x - \sin x)(\cos x + \sin x)$:$L = \lim _{x \rightarrow \frac{\pi}{4}} \frac{5(\cos x+\sin x)^4(\cos x-\sin x)}{2(\cos x-\sin x)(\cos x+\sin x)}$.$L = \lim _{x \rightarrow \frac{\pi}{4}} \frac{5}{2}(\cos x+\sin x)^3$.Substituting $x = \frac{\pi}{4}$:$L = \frac{5}{2}(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}})^3 = \frac{5}{2}(\sqrt{2})^3 = \frac{5}{2}(2 \sqrt{2}) = 5 \sqrt{2}$.

$\lim _{x \rightarrow \frac{\pi}{4}} \frac{4 \sqrt{2}-(\cos x+\sin x)^5}{1-\sin 2 x} = $

Similar Questions

$\mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + x)}^{1/2}} - {{(1 - x)}^{1/2}}}}{x} = $

If $f(x)$ is a differentiable function and $f''(0) = a$,then $\mathop {\lim }\limits_{x \to 0} \frac{2f(x) - 3f(2x) + f(4x)}{x^2}$ is (in $a$)

The value of the limit of $\frac{x^3 - x^2 - 18}{x - 3}$ as $x$ tends to $3$ is

If $f(a) = 2$,$f'(a) = 1$,$g(a) = -3$,$g'(a) = -1$,then $\mathop {\lim }\limits_{x \to a} \,\frac{f(a)g(x) - f(x)g(a)}{x - a} = $

$\mathop {\lim }\limits_{x \to a} \frac{{\sqrt {3x - a} - \sqrt {x + a} }}{{x - a}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module