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Question #40354

Difficulty: 8/10 | 🟢 Free

The Question

Let $f(x) = \frac{x e^{x-1} - 1}{|x-1| + \sin(x-1)}$ for $x \neq 1$ . If $L_1 = \lim_{x \to 1^+} f(x)$ and $L_2 = \lim_{x \to 1^-} f(x)$ , then which of the following is correct?

Options

A.
$L_1 = 1, L_2 = -\infty$
B.
$L_1 = 1, L_2 = \infty$
C.
$L_1 = \frac{1}{2}, L_2 = -\infty$
D.
$L_1 = 1, L_2 = 1$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Rational Function Asymptotes|Indeterminate Forms|Geometrical Interpretation of Modulus

Why This Question Is Tricky

Common Mistakes

This question tests the evaluation of indeterminate forms where the denominator's leading terms cancel asymmetrically due to a modulus function.

The primary trap lies in the left-hand limit: students often reflexively assume $|x-1| + \sin(x-1)$ behaves symmetrically or incorrectly apply L'Hôpital's rule without resolving the modulus sign.

By forcing the first-order terms in the denominator to cancel exactly ( $h - \sin h \approx h^3/6$ ), the denominator drops to a higher-order zero.

This creates a vertical asymptote ( $-\infty$ ) on the left, while the right side remains a finite removable discontinuity, subverting the expectation of a standard $0/0$ limit.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Relations and Functions - Class 11
Concept	Rational Function Asymptotes
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free