Question:
\(\displaystyle\lim_{x \to \pi/4} \frac{\cot^3 x - \tan x}{\cos\left(x + \frac{\pi}{4}\right)}\) is :
\(\displaystyle\lim_{x \to \pi/4} \frac{\cot^3 x - \tan x}{\cos\left(x + \frac{\pi}{4}\right)}\) is :
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The limit is of the form \(0/0\); apply L'Hospital's rule, or factor in terms of \(t=\tan x\).
Updated On: Nov 5, 2026
- \(8\sqrt{2}\)
- \(4\)
- \(4\sqrt{2}\)
- \(8\)
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The Correct Option is D
Solution and Explanation
At \(x = \frac\pi4\): numerator \(= 1 - 1 = 0\), denominator \(= \cos\frac\pi2 = 0\), so L'Hospital's rule applies.
Derivative of numerator: \(-3\cot^2 x\csc^2 x - \sec^2 x\); at \(x = \frac\pi4\): \(-3(1)(2) - 2 = -8\).
Derivative of denominator: \(-\sin\left(x + \frac\pi4\right) = -1\) at \(x = \frac\pi4\).
Limit \(= \frac{-8}{-1} = 8\).
Answer: (4) 8
Derivative of numerator: \(-3\cot^2 x\csc^2 x - \sec^2 x\); at \(x = \frac\pi4\): \(-3(1)(2) - 2 = -8\).
Derivative of denominator: \(-\sin\left(x + \frac\pi4\right) = -1\) at \(x = \frac\pi4\).
Limit \(= \frac{-8}{-1} = 8\).
Answer: (4) 8
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