\( \lim_{x \to 0} \frac{e^{2|\sin x|} - 2|\sin x| - 1}{x^2} \)
- is equal to \( -1 \)
- does not exist
- is equal to \( 1 \)
- is equal to \( 2 \)
The Correct Option is D
Solution and Explanation
To evaluate the limit \(\lim_{x \to 0} \frac{e^{2|\sin x|} - 2|\sin x| - 1}{x^2}\), we follow these steps:
For values of \(x\) close to zero, \(|\sin x| \approx |x|\).
Using the Taylor expansion for \(e^y\) around \(y=0\), which is \(e^y \approx 1 + y + \frac{y^2}{2} + \cdots\), we substitute \(y = 2|\sin x|\):
\(e^{2|\sin x|} \approx 1 + 2|\sin x| + \frac{(2|\sin x|)^2}{2} = 1 + 2|\sin x| + 2|\sin x|^2\)
Substitute this approximation into the limit expression:
\(\frac{(1 + 2|\sin x| + 2|\sin x|^2) - 2|\sin x| - 1}{x^2}\)
Simplify the numerator:
\(\frac{2|\sin x|^2}{x^2}\)
Since \(|\sin x|^2 \approx x^2\) as \(x \to 0\), the expression becomes:
\(\frac{2x^2}{x^2} = 2\)
Therefore, the limit is:
\(\lim_{x \to 0} \frac{e^{2|\sin x|} - 2|\sin x| - 1}{x^2} = 2\)
The result is \(2\).
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