Question

Which one of the following is the CORRECT value of \(y\), as defined by the expression given below?
\[ y = \lim_{x \to 0} \frac{2x}{e^x - 1} \]

• A. 1
• B. 2
• C. 0
• D. \(\infty\)

Hint:

- For limits of the type \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), L'Hôpital's Rule is a powerful tool.
- Alternatively, use series expansion: \(e^x - 1 \approx x\) when \(x \to 0\), giving \(\frac{2x}{x} = 2\).

Answer 1

The Correct Option is B

Step 1: As \(x \to 0\), numerator \(2x \to 0\) and denominator \(e^x - 1 \to 0\). Thus, the limit is of the form \(\frac{0}{0}\), so L'Hôpital's Rule can be applied. Step 2: Differentiate numerator and denominator: \[ \lim_{x \to 0} \frac{2x}{e^x - 1} = \lim_{x \to 0} \frac{(2)'}{(e^x - 1)'} = \lim_{x \to 0} \frac{2}{e^x}. \] Step 3: Evaluate at \(x=0\): \[ \frac{2}{e^0} = \frac{2}{1} = 2. \] \[ \boxed{2} \]