Key Concepts and Formulas

• Standard Limit: The fundamental limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$.
• Taylor Series Expansion: The Taylor series expansion of $e^u$ around $u=0$ is $e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \dots$.
• L'Hôpital's Rule: If $\lim_{x \to c} f(x) = 0$ and $\lim_{x \to c} g(x) = 0$, then $\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$, provided the latter limit exists.
• Properties of Limits: The limit of a product is the product of the limits, i.e., $\lim_{x \to c} [f(x)g(x)] = [\lim_{x \to c} f(x)][\lim_{x \to c} g(x)]$, provided both limits exist.

Step-by-Step Solution

Step 1: Analyze the expression and identify indeterminate forms. The given limit is $\lim _{x \rightarrow 0} \frac{e^{2|\sin x|}-2|\sin x|-1}{x^2}$. As $x \rightarrow 0$, $|\sin x| \rightarrow 0$. The numerator approaches $e^{2(0)} - 2(0) - 1 = e^0 - 0 - 1 = 1 - 1 = 0$. The denominator approaches $0^2 = 0$. This is an indeterminate form of type $\frac{0}{0}$, suggesting that we can use L'Hôpital's Rule or Taylor series expansion.

Step 2: Manipulate the expression to utilize known limits. We can rewrite the expression by multiplying and dividing by $|\sin x|^2$, which is equivalent to $\sin^2 x$ as $x \to 0$. This is done to isolate a standard limit form. $\lim _{x \rightarrow 0} \frac{e^{2|\sin x|}-2|\sin x|-1}{x^2} = \lim _{x \rightarrow 0} \frac{e^{2|\sin x|}-2|\sin x|-1}{|\sin x|^2} \times \frac{|\sin x|^2}{x^2}$ We know that $\lim_{x \to 0} \frac{\sin x}{x} = 1$, which implies $\lim_{x \to 0} \frac{\sin^2 x}{x^2} = \left(\lim_{x \to 0} \frac{\sin x}{x}\right)^2 = 1^2 = 1$. Since $|\sin x|^2 = \sin^2 x$ for all $x$, we have $\lim _{x \rightarrow 0} \frac{|\sin x|^2}{x^2} = \lim _{x \rightarrow 0} \frac{\sin ^2 x}{x^2} = 1$.

Step 3: Evaluate the first part of the limit by substitution. Let $t = |\sin x|$. As $x \rightarrow 0$, we have $t \rightarrow 0$. The first part of the limit becomes: $\lim _{x \rightarrow 0} \frac{e^{2|\sin x|}-2|\sin x|-1}{|\sin x|^2} = \lim _{t \rightarrow 0} \frac{e^{2t}-2t-1}{t^2}$ This is still an indeterminate form of type $\frac{0}{0}$.

Step 4: Apply L'Hôpital's Rule to the transformed limit. We apply L'Hôpital's Rule to $\lim _{t \rightarrow 0} \frac{e^{2t}-2t-1}{t^2}$. The derivative of the numerator with respect to $t$ is $\frac{d}{dt}(e^{2t}-2t-1) = 2e^{2t} - 2$. The derivative of the denominator with respect to $t$ is $\frac{d}{dt}(t^2) = 2t$. So, the limit becomes: $\lim _{t \rightarrow 0} \frac{2e^{2t}-2}{2t}$ This is still an indeterminate form of type $\frac{0}{0}$ ($2e^0 - 2 = 0$ and $2(0) = 0$).

Step 5: Apply L'Hôpital's Rule a second time. We apply L'Hôpital's Rule again to $\lim _{t \rightarrow 0} \frac{2e^{2t}-2}{2t}$. The derivative of the numerator with respect to $t$ is $\frac{d}{dt}(2e^{2t}-2) = 2(2e^{2t}) = 4e^{2t}$. The derivative of the denominator with respect to $t$ is $\frac{d}{dt}(2t) = 2$. So, the limit becomes: $\lim _{t \rightarrow 0} \frac{4e^{2t}}{2} = \frac{4e^{2(0)}}{2} = \frac{4e^0}{2} = \frac{4 \times 1}{2} = 2$

Step 6: Combine the results of the two parts of the limit. We have decomposed the original limit into a product of two limits: $\lim _{x \rightarrow 0} \frac{e^{2|\sin x|}-2|\sin x|-1}{x^2} = \left(\lim _{t \rightarrow 0} \frac{e^{2t}-2t-1}{t^2}\right) \times \left(\lim _{x \rightarrow 0} \frac{\sin ^2 x}{x^2}\right)$ From Step 3 and Step 5, we found $\lim _{t \rightarrow 0} \frac{e^{2t}-2t-1}{t^2} = 2$. From Step 2, we found $\lim _{x \rightarrow 0} \frac{\sin ^2 x}{x^2} = 1$. Therefore, the original limit is $2 \times 1 = 2$.

Alternative Approach using Taylor Series:

Step A1: Use the Taylor series expansion for $e^u$. For small $u$, $e^u = 1 + u + \frac{u^2}{2!} + O(u^3)$. Let $u = 2|\sin x|$. As $x \to 0$, $|\sin x| \to 0$, so $u \to 0$. We also know that for small $x$, $\sin x = x - \frac{x^3}{3!} + O(x^5)$. Therefore, $|\sin x| = |x - \frac{x^3}{6} + \dots| = |x|(1 - \frac{x^2}{6} + \dots)$. For $x \to 0$, $|\sin x| \approx |x|$. More precisely, $|\sin x| = |x| + O(x^3)$. Thus, $u = 2|\sin x| = 2|x| + O(x^3)$.

Step A2: Substitute the Taylor expansion into the numerator. $e^{2|\sin x|} = 1 + (2|\sin x|) + \frac{(2|\sin x|)^2}{2!} + O((|\sin x|)^3)$ $e^{2|\sin x|} = 1 + 2|\sin x| + 2|\sin x|^2 + O(|\sin x|^3)$ Since $|\sin x|^2 = \sin^2 x$, $e^{2|\sin x|} = 1 + 2|\sin x| + 2\sin^2 x + O(x^3)$ (since $|\sin x|^3$ is of order $x^3$).

Step A3: Substitute this back into the original expression. The numerator is $e^{2|\sin x|}-2|\sin x|-1$. Numerator = $(1 + 2|\sin x| + 2\sin^2 x + O(x^3)) - 2|\sin x| - 1$ Numerator = $2\sin^2 x + O(x^3)$.

Step A4: Evaluate the limit. $\lim _{x \rightarrow 0} \frac{2\sin^2 x + O(x^3)}{x^2} = \lim _{x \rightarrow 0} \left(\frac{2\sin^2 x}{x^2} + \frac{O(x^3)}{x^2}\right)$ $= \lim _{x \rightarrow 0} \frac{2\sin^2 x}{x^2} + \lim _{x \rightarrow 0} O(x)$ $= 2 \lim _{x \rightarrow 0} \left(\frac{\sin x}{x}\right)^2 + 0$ $= 2 (1)^2 = 2$

Both methods yield the same result.

Common Mistakes & Tips

• Incorrectly handling absolute values: Remember that $|\sin x|^2 = \sin^2 x$, which simplifies calculations. When using Taylor series, be mindful of the behavior of $|\sin x|$ near zero.
• Errors in differentiation for L'Hôpital's Rule: Double-check the derivatives of exponential functions and trigonometric functions. Ensure that the conditions for L'Hôpital's Rule are met at each application.
• Forgetting the standard limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$: This limit is crucial for simplifying parts of the expression and is often combined with Taylor series expansions or L'Hôpital's Rule.
• Taylor expansion of $e^u$: Use the expansion $e^u = 1 + u + \frac{u^2}{2!} + \dots$. When substituting $u = 2|\sin x|$, be careful with the order of the terms. For $x \to 0$, $|\sin x| \sim |x|$, so $(2|\sin x|)^2 \sim 4x^2$.

Summary

The problem requires evaluating a limit that results in an indeterminate form $\frac{0}{0}$. We can solve this by either applying L'Hôpital's Rule multiple times after algebraic manipulation or by using the Taylor series expansion of the exponential function. Both methods involve recognizing that as $x \to 0$, $|\sin x| \to 0$ and $\frac{\sin x}{x} \to 1$. The manipulation involves separating the expression into a part that resembles the Taylor series of $e^u$ and a part that resembles the standard limit $\frac{\sin x}{x}$. Using L'Hôpital's Rule twice on the transformed expression or using Taylor expansions leads to the evaluation of the limit as 2.

The final answer is $\boxed{2}$.