Question

If for a suitable \( \alpha>0 \), \[ \lim_{x \to 0} \left( \frac{1}{e^{2x} - 1} - \frac{1}{\alpha x} \right) \] exists and is equal to \( l \) (\( |l|<\infty \)), then \( \alpha = 2, l = -\frac{1}{2} \) is given by

• A. \( \alpha = 2, l = 2 \)
• B. \( \alpha = 2, l = -\frac{1}{2} \)
• C. \( \alpha = \frac{1}{2}, l = -2 \)
• D. \( \alpha = \frac{1}{2}, l = \frac{1}{2} \)

Hint:

When dealing with limits involving series or exponential functions, use series expansions to simplify the expressions and find the correct limits.

Answer 1

The Correct Option is B

Step 1: Analyze the limit expression.
We need to evaluate the limit of the expression as \( x \to 0 \). Begin by expanding the terms and simplifying the expression to find the correct value of \( \alpha \) and \( l \).
Step 2: Find \( \alpha \) and \( l \).
Using series expansions for \( e^{2x} \), we find that \( \alpha = 2 \) and \( l = -\frac{1}{2} \).