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Limits Continuity and Differentiability question

2019 · 12 Jan · Shift 1 · Q39

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Limits Continuity and Differentiability
2019 · 12 Jan · Shift 1 · Q39

Limits Continuity and Differentiability question

2019 · 12 Jan · Shift 1 · Q39

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1

\mathop {\lim }\limits_{x \to \pi /4} {{{{\cot }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}}

is :

A
$8\sqrt 2$
B
4
C
$4\sqrt 2$
D
8

View written solutionFree

Correct answer: D

We need to evaluate $L=\lim_{x\to \pi/4}\frac{\cot^3 x-\tan x}{\cos\left(x+\frac\pi4\right)}.$
First check the form at $x=\frac\pi4$ :

$\tan\frac\pi4=1$
$\cot\frac\pi4=1$
so numerator $=1^3-1=0$
denominator $=\cos\left(\frac\pi4+\frac\pi4\right)=\cos\frac\pi2=0$

So the limit is of the form $\frac00$ , and we can simplify.

Rewrite the numerator: $\cot^3 x-\tan x=\frac{\cos^3 x}{\sin^3 x}-\frac{\sin x}{\cos x}.$ Taking LCM $\sin^3 x\cos x$ , $\cot^3 x-\tan x=\frac{\cos^4 x-\sin^4 x}{\sin^3 x\cos x}.$ Now, $\cos^4 x-\sin^4 x=(\cos^2 x-\sin^2 x)(\cos^2 x+\sin^2 x)=\cos 2x.$ Hence, $\cot^3 x-\tan x=\frac{\cos 2x}{\sin^3 x\cos x}.$
Also simplify the denominator:

=\frac{\cos x-\sin x}{\sqrt2}.$$ So, $$L=\lim_{x\to \pi/4}\frac{\frac{\cos 2x}{\sin^3 x\cos x}}{\frac{\cos x-\sin x}{\sqrt2}}.$$ Using $$\cos 2x=(\cos x-\sin x)(\cos x+\sin x),$$ we get $$L=\lim_{x\to \pi/4}\frac{\sqrt2(\cos x-\sin x)(\cos x+\sin x)}{\sin^3 x\cos x(\cos x-\sin x)}.$$ Cancel $(\cos x-\sin x)$: $$L=\lim_{x\to \pi/4}\frac{\sqrt2(\cos x+\sin x)}{\sin^3 x\cos x}.$$ 5. Now substitute $x=\frac\pi4$: $$\sin\frac\pi4=\cos\frac\pi4=\frac1{\sqrt2}.$$ Then $$\cos\frac\pi4+\sin\frac\pi4=\frac1{\sqrt2}+\frac1{\sqrt2}=\sqrt2.$$ Also, $$\sin^3\frac\pi4\cdot \cos\frac\pi4=\left(\frac1{\sqrt2}\right)^3\left(\frac1{\sqrt2}\right)=\left(\frac1{\sqrt2}\right)^4=\frac14.$$ Therefore, $$L=\frac{\sqrt2\cdot \sqrt2}{1/4}=\frac{2}{1/4}=8.$$ 6. Hence the correct option is $$\boxed{8}.$$ That corresponds to **Option D**.

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