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Question #40923

Difficulty: 8/10 | 🟢 Free

The Question

Let $f(x) = \sqrt{x^2 - 1 - 2ix} + \sqrt{x^2 - 1 + 2ix}$ , where $\sqrt{z}$ denotes the principal square root of $z$ (i.e., $-\pi < \text{Arg}(z) \le \pi$ ). Evaluate the limit: $L = \lim_{x \to 0^-} \left( \cos x + f(x) \right)^{\frac{1}{\sin x}}$

Options

A.
$e^{-2}$
B.
$e^{2}$
C.
$e^{-1}$
D.
$1$

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

Definition of Limit|De Moivre’s Theorem Applications|Algebraic Manipulation of i|Square Root and Modulus Logic

Why This Question Is Tricky

Common Mistakes

This question tests the lethal trap of blindly simplifying the principal square root of a complex square, $\sqrt{w^2} = w$ .

Students commonly assume $f(x) = (x-i) + (x+i) = 2x$ , which leads to the incorrect limit $e^2$ .

However, the principal branch of the square root requires the argument to be in $(-\pi/2, \pi/2]$ .

For $x < 0$ , the complex numbers $x \pm i$ have negative real parts, meaning they lie outside this principal domain.

Thus, the correct simplification is $\sqrt{(x \pm i)^2} = -(x \pm i)$ , yielding $f(x) = -2x$ .

This elegantly combines complex branch cuts with $1^\infty$ indeterminate limit evaluation, forcing students to account for the discontinuity of the argument function at the negative real axis.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Limits and Derivatives - Class 11
Concept	Definition of Limit
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free