Question

\( \lim_{x \to 0} \frac{e^{2|\sin x|} - 2|\sin x| - 1}{x^2} \)

• A. is equal to \( -1 \)
• B. does not exist
• C. is equal to \( 1 \)
• D. is equal to \( 2 \)

Answer 1

The Correct Option is D

Consider the limit:

\[ \lim_{x \to 0} \frac{e^{2|\sin x|} - 2|\sin x| - 1}{x^2} \]

Expanding \( e^{2|\sin x|} \) around \( x = 0 \):

\[ \lim_{x \to 0} \frac{e^{2|\sin x|} - 2|\sin x| - 1}{|\sin x|^2} \cdot \frac{\sin^2 x}{x^2} \]

Let \( |\sin x| = t \):

\[ \lim_{t \to 0} \frac{e^{2t} - 2t - 1}{t^2} \cdot \lim_{x \to 0} \frac{\sin^2 x}{x^2} \]

Evaluating the first limit:

\[ \lim_{t \to 0} \frac{2e^{2t} - 2}{2t} \cdot 1 = 2 \times 1 = 2 \]