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Question #41306

Difficulty: 8/10 | 🟢 Free

The Question

Evaluate the limit: lim⁡x→π4+2−2sin⁡2x+2(cos⁡x−sin⁡x)+2ln⁡(1+tan⁡(x−π4))cos⁡2x\lim_{x \to \frac{\pi}{4}^+} \frac{\sqrt{2 - 2\sin 2x} + \sqrt{2}(\cos x - \sin x) + 2\ln\left(1 + \tan\left(x - \frac{\pi}{4}\right)\right)}{\cos 2x}

Options

  • A.

    -1

  • B.

    1

  • C.

    -3

  • D.

    0

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

The Zero-Factor Cancellation Trap|Indeterminate Forms|Square Root and Modulus Logic

Why This Question Is Tricky

Common Mistakes

This question tests the 'Zero-Factor Cancellation Trap' combined with modulus logic.

Students frequently make the error of simplifying f(x)2\sqrt{f(x)^2} to f(x)f(x) instead of ∣f(x)∣|f(x)|.

In this problem, assuming (cos⁡x−sin⁡x)2=cos⁡x−sin⁡x\sqrt{(\cos x - \sin x)^2} = \cos x - \sin x leads to a non-zero trigonometric term 22(cos⁡x−sin⁡x)2\sqrt{2}(\cos x - \sin x) in the numerator.

When divided by the zero-factor cos⁡2x=(cos⁡x−sin⁡x)(cos⁡x+sin⁡x)\cos 2x = (\cos x - \sin x)(\cos x + \sin x), this incorrectly cancels out to yield a false limit component of +2+2, leading to the trap answer of 11.

Recognizing the right-hand limit direction correctly evaluates the modulus to sin⁡x−cos⁡x\sin x - \cos x, causing the zero-factors to perfectly annihilate before any indeterminate evaluation is needed.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterContinuity and Differentiability - Class 12
ConceptThe Zero-Factor Cancellation Trap
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree