Question

Find the value of the following limit:
\[ \lim_{x \to 0} \frac{e^{x} - 1}{x^2} \]

• A. 1
• B. 0
• C. \( \frac{1}{2} \)
• D. 2

Hint:

In cases of indeterminate forms like \( \frac{0}{0} \), use L'Hopital's Rule, which involves differentiating the numerator and denominator.

Answer 1

The Correct Option is A

We are given the limit: \[ \lim_{x \to 0} \frac{e^{x} - 1}{x^2}. \] Step 1: Apply L'Hopital's Rule This is an indeterminate form \( \frac{0}{0} \), so we can apply L'Hopital's Rule, which involves differentiating the numerator and denominator.
First, differentiate the numerator: \[ \frac{d}{dx}(e^{x} - 1) = e^{x}. \] Now, differentiate the denominator: \[ \frac{d}{dx}(x^2) = 2x. \] Now, we can compute the new limit: \[ \lim_{x \to 0} \frac{e^{x}}{2x}. \] Step 2: Apply L'Hopital's Rule Again This is again in the form \( \frac{0}{0} \), so we apply L'Hopital's Rule once more.
Differentiate the numerator: \[ \frac{d}{dx}(e^{x}) = e^{x}. \] Differentiate the denominator: \[ \frac{d}{dx}(2x) = 2. \] Now, the limit becomes: \[ \lim_{x \to 0} \frac{e^{x}}{2} = \frac{1}{2}. \] Thus, the value of the limit is (C) \( \frac{1}{2} \).