\( \lim_{x \to 0} \frac{e^{2|\sin x|} - 2|\sin x| - 1}{x^2} \)
- is equal to \( -1 \)
- does not exist
- is equal to \( 1 \)
- is equal to \( 2 \)
The Correct Option is D
Solution and Explanation
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:
\[ \lim_{t \to 0} \frac{2e^{2t} - 2}{2t} \cdot 1 = 2 \times 1 = 2 \]
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