Question

The value of \( \lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{1 - \cos(x)} \) is equal to:

• A. 2
• B. 1
• C. 0
• D. -1

Hint:

When a limit results in the indeterminate form \( \frac{0}{0} \), apply L'Hopital's Rule by differentiating the numerator and denominator.

Answer 1

The Correct Option is C

Step 1: Apply L'Hopital's Rule. The given limit is of the indeterminate form \( \frac{0}{0} \). We can apply L'Hopital's Rule, which requires us to differentiate the numerator and the denominator. The numerator is: \[ f(x) = e^x - e^{-x} - 2x \quad \Rightarrow \quad f'(x) = e^x + e^{-x} - 2. \] The denominator is: \[ g(x) = 1 - \cos(x) \quad \Rightarrow \quad g'(x) = \sin(x). \] Step 2: Evaluate the limit. Now, compute the limit: \[ \lim_{x \to 0} \frac{e^x + e^{-x} - 2}{\sin(x)} = \frac{2 - 2}{0} = 0. \] Thus, the correct answer is: \[ \boxed{0}. \]