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

To solve the given limit problem, we need to evaluate \(\lim_{x \to 0} \frac{e^{2|\sin x|} - 2|\sin x| - 1}{x^2}\). Let's break down the solution into clear steps:

For small values of \(x\), we know that \(|\sin x| \approx |x|\). Therefore, \(|\sin x|\) behaves like \(|x|\) near zero. 

The expression \(e^{2|\sin x|}\) can be expanded using the Taylor series for \(e^y\) around \(y=0\): \(e^y \approx 1 + y + \frac{y^2}{2} + \cdots\). Applying this to \(2|\sin x|\), we get:

\(e^{2|\sin x|} \approx 1 + 2|\sin x| + 2|\sin x|^2/2 + \cdots = 1 + 2|\sin x| + 2|\sin x|^2\)

Substituting the approximations in the original limit expression, we have:

\(\frac{e^{2|\sin x|} - 2|\sin x| - 1}{x^2} \approx \frac{(1 + 2|\sin x| + 2|\sin x|^2) - 2|\sin x| - 1}{x^2}\)

Simplifying this, we get:

\(\frac{2|\sin x|^2}{x^2}\)

Knowing \(|\sin x|^2 \approx x^2\) when \(x \to 0\), the expression becomes:

\(\frac{2x^2}{x^2} = 2\)

Thus, by evaluating the above expression, we arrive at:

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

This confirms that the correct answer is \(2\).

Answer 2

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:

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