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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Limits

NTA Abhyas 2020 · Mathematics · Limits

If L = _ x → (π)/(4) (1 - x)(1 - 2x) (1 + x) (π - 4x) ^ 3 , then...

If L=lim⁡x→π4(1−tan⁡x)(1−sin⁡2x)(1+tan⁡x)(π−4x)3L = \lim_{x \to \dfrac{\pi}{4}}\dfrac{\left(1 - \tan x\right)\left(1 - \sin 2x\right)}{\left(1 + \tan x\right){\left(\pi - 4x\right)}^{3}}, then the value of 40L40L is equal to

Correct answer

1.25

Step-by-step solution

L = \lim_{x \to \dfrac{\pi}{4}}\dfrac{\text\tan\left(\dfrac{\pi}{4} - x\right)\left(1 - \sin 2x\right)}{4\left(\dfrac{\pi}{4} - x\right){\left(\pi - 4x\right)}^{2}}
=lim⁡x→π414(1−sin⁡2x)(π−4x)2= \lim_{x \to \dfrac{\pi}{4}}\dfrac{1}{4}\dfrac{\left(1 - \sin 2x\right)}{{\left(\pi - 4x\right)}^{2}}
=lim⁡x→π414(−cos⁡2x)⋅22(π−4x)(−4)= \lim_{x \to \dfrac{\pi}{4}}\dfrac{1}{4}\dfrac{\left(- \cos 2x\right) ⋅ 2}{2\left(\pi - 4x\right)\left(- 4\right)} (applying L-Hospital rule)
=lim⁡x→π4116cos⁡(2x)(π−4x)= \lim_{x \to \dfrac{\pi}{4}}\dfrac{1}{16}\dfrac{\cos \left(2x\right)}{\left(\pi - 4x\right)}
=lim⁡x→π4116(−sin⁡2x)×2−4= \lim_{x \to \dfrac{\pi}{4}}\dfrac{1}{16}\dfrac{\left(- \sin 2x\right) \times 2}{- 4} (again applying L-Hospital rule)
=116(−1−4)×2⇒L=132= \dfrac{1}{16}\left(\dfrac{- 1}{- 4}\right) \times 2 ⇒ L = \dfrac{1}{32}
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