The value of mathop {lim }limits_{x to pi /4} | vedclass

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Question

(D) Let $f(x) = \cot^3 x - 	an x$ and $g(x) = \cos(x + \pi/4)$.At $x = \pi/4$,$f(\pi/4) = 1^3 - 1 = 0$ and $g(\pi/4) = \cos(\pi/2) = 0$. This is a $0

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) Let $f(x) = \cot^3 x - 	an x$ and $g(x) = \cos(x + \pi/4)$.At $x = \pi/4$,$f(\pi/4) = 1^3 - 1 = 0$ and $g(\pi/4) = \cos(\pi/2) = 0$. This is a $0/0$ form.Applying $L$'$H$ôpital's rule,we differentiate the numerator and denominator with respect to $x$:Numerator derivative: $\frac{d}{dx}(\cot^3 x - 	an x) = 3\cot^2 x(-\csc^2 x) - \sec^2 x$.Denominator derivative: $\frac{d}{dx}(\cos(x + \pi/4)) = -\sin(x + \pi/4)$.Now,evaluate the limit as $x 	o \pi/4$:$\lim_{x 	o \pi/4} \frac{-3\cot^2 x \csc^2 x - \sec^2 x}{-\sin(x + \pi/4)} = \frac{-3(1)^2(\sqrt{2})^2 - (\sqrt{2})^2}{-\sin(\pi/2)} = \frac{-3(2) - 2}{-1} = \frac{-8}{-1} = 8$.

The value of $\mathop {\lim }\limits_{x \to \pi /4} \frac{{{{\cot }^3}x - \tan x}}{{\cos \left( {x + \pi /4} \right)}}$ is

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