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Limits Continuity and Differentiability question

2017 · Shift 0 · Q43
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  5. /2017 · Shift 0 · Q43

Limits Continuity and Differentiability question

2017 · Shift 0 · Q43

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
lim⁡x→π2cot⁡x−cos⁡x(π−2x)3\mathop {\lim }\limits_{x \to {\pi \over 2}} {{\cot x - \cos x} \over {{{\left( {\pi - 2x} \right)}^3}}}x→2π​lim​(π−2x)3cotx−cosx​ equals
  1. A
    116{1 \over {16}}161​
  2. B
    18{1 \over 8}81​
  3. C
    14{1 \over {4}}41​
  4. D
    124{1 \over {24}}241​
View written solutionFree

Correct answer: A

  1. We need to evaluate
L=lim⁡x→π2cot⁡x−cos⁡x(π−2x)3.L=\lim_{x\to \frac\pi2}\frac{\cot x-\cos x}{(\pi-2x)^3}.L=x→2π​lim​(π−2x)3cotx−cosx​.
  1. Rewrite the denominator:
π−2x=−2(x−π2).\pi-2x=-2\left(x-\frac\pi2\right).π−2x=−2(x−2π​).

So

(π−2x)3=−8(x−π2)3.(\pi-2x)^3=-8\left(x-\frac\pi2\right)^3.(π−2x)3=−8(x−2π​)3.

Hence

L=−18lim⁡x→π2cot⁡x−cos⁡x(x−π2)3.L=-\frac18\lim_{x\to \frac\pi2}\frac{\cot x-\cos x}{\left(x-\frac\pi2\right)^3}.L=−81​x→2π​lim​(x−2π​)3cotx−cosx​.
  1. Put
h=x−π2⇒x=π2+h,h=x-\frac\pi2 \quad \Rightarrow \quad x=\frac\pi2+h,h=x−2π​⇒x=2π​+h,

with h→0h\to 0h→0. Then

cos⁡x=cos⁡(π2+h)=−sin⁡h,\cos x=\cos\left(\frac\pi2+h\right)=-\sin h,cosx=cos(2π​+h)=−sinh,

and

cot⁡x=cot⁡(π2+h)=−tan⁡h.\cot x=\cot\left(\frac\pi2+h\right)=-\tan h.cotx=cot(2π​+h)=−tanh.

Therefore,

cot⁡x−cos⁡x=(−tan⁡h)−(−sin⁡h)=sin⁡h−tan⁡h.\cot x-\cos x=(-\tan h)-(-\sin h)=\sin h-\tan h.cotx−cosx=(−tanh)−(−sinh)=sinh−tanh.

Also,

π−2x=π−2(π2+h)=−2h,\pi-2x=\pi-2\left(\frac\pi2+h\right)=-2h,π−2x=π−2(2π​+h)=−2h,

so

L=lim⁡h→0sin⁡h−tan⁡h(−2h)3=−18lim⁡h→0sin⁡h−tan⁡hh3.L=\lim_{h\to 0}\frac{\sin h-\tan h}{(-2h)^3} = -\frac18\lim_{h\to 0}\frac{\sin h-\tan h}{h^3}.L=h→0lim​(−2h)3sinh−tanh​=−81​h→0lim​h3sinh−tanh​.
  1. Simplify the numerator:
sin⁡h−tan⁡h=sin⁡h−sin⁡hcos⁡h=sin⁡h(1−1cos⁡h)=sin⁡h⋅cos⁡h−1cos⁡h.\sin h-\tan h=\sin h-\frac{\sin h}{\cos h} =\sin h\left(1-\frac1{\cos h}\right) =\sin h\cdot \frac{\cos h-1}{\cos h}.sinh−tanh=sinh−coshsinh​=sinh(1−cosh1​)=sinh⋅coshcosh−1​.

Thus,

sin⁡h−tan⁡hh3=sin⁡hh⋅cos⁡h−1h2⋅1cos⁡h.\frac{\sin h-\tan h}{h^3} =\frac{\sin h}{h}\cdot \frac{\cos h-1}{h^2}\cdot \frac1{\cos h}.h3sinh−tanh​=hsinh​⋅h2cosh−1​⋅cosh1​.

Taking limit as h→0h\to 0h→0:

  • sin⁡hh→1\displaystyle \frac{\sin h}{h}\to 1hsinh​→1
  • cos⁡h−1h2→−12\displaystyle \frac{\cos h-1}{h^2}\to -\frac12h2cosh−1​→−21​
  • 1cos⁡h→1\displaystyle \frac1{\cos h}\to 1cosh1​→1

Hence,

lim⁡h→0sin⁡h−tan⁡hh3=1⋅(−12)⋅1=−12.\lim_{h\to 0}\frac{\sin h-\tan h}{h^3}=1\cdot \left(-\frac12\right)\cdot 1=-\frac12.h→0lim​h3sinh−tanh​=1⋅(−21​)⋅1=−21​.

Therefore,

L=−18(−12)=116.L=-\frac18\left(-\frac12\right)=\frac1{16}.L=−81​(−21​)=161​.
  1. So the correct option is
116.\boxed{\frac1{16}}.161​​.

This is option A\boxed{A}A​.

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