Evaluate a Limit Involving Exponentials and Trigonometric Functions

To evaluate the limit $\lim_{x \to 0} \frac{e^{2x} - 1 - 2x}{x^2 \sin x}$, we can use Taylor series expansions for the functions involved around $x=0$. The Taylor series expansion for $e^u$ around $u=0$ is $e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \dots$. Substituting $u=2x$, we get $e^{2x} = 1 + (2x) + \frac{(2x)^2}{2!} + \frac{(2x)^3}{3!} + \dots = 1 + 2x + \frac{4x^2}{2} + \frac{8x^3}{6} + \dots = 1 + 2x + 2x^2 + \frac{4}{3}x^3 + \dots$. So, the numerator $e^{2x} - 1 - 2x$ becomes: $(1 + 2x + 2x^2 + \frac{4}{3}x^3 + \dots) - 1 - 2x = 2x^2 + \frac{4}{3}x^3 + \dots$. The Taylor series expansion for $\sin x$ around $x=0$ is $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots = x - \frac{x^3}{6} + \dots$. So, the denominator $x^2 \sin x$ becomes: $x^2 (x - \frac{x^3}{6} + \dots) = x^3 - \frac{x^5}{6} + \dots$. Now, we can rewrite the limit expression using these expansions: $$ \lim_{x \to 0} \frac{2x^2 + \frac{4}{3}x^3 + \dots}{x^3 - \frac{x^5}{6} + \dots} $$ To evaluate this limit, we can divide both the numerator and the denominator by the lowest power of $x$ present in the denominator, which is $x^3$. However, it is more direct to factor out the lowest power of $x$ from both the numerator and the denominator: $$ \lim_{x \to 0} \frac{x^2(2 + \frac{4}{3}x + \dots)}{x^3(1 - \frac{x^2}{6} + \dots)} $$ This simplifies to: $$ \lim_{x \to 0} \frac{2 + \frac{4}{3}x + \dots}{x(1 - \frac{x^2}{6} + \dots)} $$ As $x$ approaches 0, the numerator approaches $2 + 0 + \dots = 2$. As $x$ approaches 0, the denominator approaches $0 imes (1 - 0 + \dots) = 0$. Since the numerator approaches a non-zero constant (2) and the denominator approaches 0, the limit will be either $\infty$ or $-\infty$. To determine the sign, we examine the behavior of the denominator as $x$ approaches 0 from the right ($x o 0^+$) and from the left ($x o 0^-$). As $x o 0^+$, the denominator $x(1 - \frac{x^2}{6} + \dots)$ approaches $0^+ imes 1 = 0^+$. Thus, the limit is $\frac{2}{0^+} = \infty$. As $x o 0^-$, the denominator $x(1 - \frac{x^2}{6} + \dots)$ approaches $0^- imes 1 = 0^-$. Thus, the limit is $\frac{2}{0^-} = -\infty$. Since the limit from the left and the limit from the right are not equal, the overall limit does not exist. However, if the question implies the limit of the absolute value or if there was a typo and the denominator was $x^3$ instead of $x^2 \sin x$, the result would be different. Let's re-examine the problem statement and ensure no steps were missed. Let's use L'Hopital's Rule as an alternative verification, as the initial form is $\frac{0}{0}$. First derivative: Numerator: $\frac{d}{dx}(e^{2x} - 1 - 2x) = 2e^{2x} - 2$ Denominator: $\frac{d}{dx}(x^2 \sin x) = 2x \sin x + x^2 \cos x$ The limit becomes $\lim_{x \to 0} \frac{2e^{2x} - 2}{2x \sin x + x^2 \cos x}$. This is still of the form $\frac{0}{0}$. Second derivative: Numerator: $\frac{d}{dx}(2e^{2x} - 2) = 4e^{2x}$ Denominator: $\frac{d}{dx}(2x \sin x + x^2 \cos x) = (2 \sin x + 2x \cos x) + (2x \cos x - x^2 \sin x) = 2 \sin x + 4x \cos x - x^2 \sin x$ The limit becomes $\lim_{x \to 0} \frac{4e^{2x}}{2 \sin x + 4x \cos x - x^2 \sin x}$. As $x o 0$, the numerator approaches $4e^0 = 4$. The denominator approaches $2 \sin 0 + 4(0)\cos 0 - 0^2 \sin 0 = 0 + 0 - 0 = 0$. Again, we have a non-zero numerator and a zero denominator. Let's analyze the denominator's sign. For small $x$, $\sin x \approx x$ and $\cos x \approx 1$. So the denominator is approximately $2x + 4x(1) - x^2(x) = 2x + 4x - x^3 = 6x - x^3 = x(6 - x^2)$. So the limit is $\lim_{x \to 0} \frac{4}{x(6 - x^2)}$. As $x o 0^+$, the denominator is $0^+(6-0) = 0^+$. The limit is $\frac{4}{0^+} = \infty$. As $x o 0^-$, the denominator is $0^-(6-0) = 0^-$. The limit is $\frac{4}{0^-} = -\infty$. Therefore, the limit does not exist. However, if the question intended for the limit to exist, there might be a misunderstanding of the prompt or a typo. Let's assume the question is precisely as stated and the limit does not exist. Let's re-evaluate the Taylor series expansion of the denominator more carefully. $\sin x = x - \frac{x^3}{6} + O(x^5)$ $x^2 \sin x = x^2 (x - \frac{x^3}{6} + O(x^5)) = x^3 - \frac{x^5}{6} + O(x^7)$ Numerator: $e^{2x} - 1 - 2x = (1 + 2x + \frac{(2x)^2}{2} + \frac{(2x)^3}{6} + O(x^4)) - 1 - 2x = (1 + 2x + 2x^2 + \frac{8x^3}{6} + O(x^4)) - 1 - 2x = 2x^2 + \frac{4}{3}x^3 + O(x^4)$. So the limit is $\lim_{x \to 0} \frac{2x^2 + \frac{4}{3}x^3 + O(x^4)}{x^3 - \frac{x^5}{6} + O(x^7)}$. Divide numerator and denominator by $x^2$: $\lim_{x \to 0} \frac{2 + \frac{4}{3}x + O(x^2)}{x - \frac{x^3}{6} + O(x^5)}$. As $x o 0$, the numerator approaches $2$. The denominator approaches $0$. Let's consider the sign of the denominator $x - \frac{x^3}{6} + O(x^5)$. For small $x$, this is dominated by the $x$ term. So, as $x o 0^+$, the denominator is positive, and as $x o 0^-$, the denominator is negative. Therefore, $\lim_{x \to 0^+} \frac{2 + \frac{4}{3}x + O(x^2)}{x - \frac{x^3}{6} + O(x^5)} = \frac{2}{0^+} = \infty$. And $\lim_{x \to 0^-} \frac{2 + \frac{4}{3}x + O(x^2)}{x - \frac{x^3}{6} + O(x^5)} = \frac{2}{0^-} = -\infty$. Since the left-hand limit and the right-hand limit are not equal, the limit does not exist. Final Answer: The limit does not exist.

To evaluate lim(x→0) [(e^(2x) - 1 - 2x) / (x^2 sin x)], I will use Taylor series expansions. Step 1: Expand e^(2x) using the Taylor series. The Taylor series for e^u around u = 0 is e^u = 1 + u + u^2/2! + u^3/3! + ... Substituting u = 2x: e^(2x) = 1 + 2x + (2x)^2/2! + (2x)^3/3! + (2x)^4/4! + ... e^(2x) = 1 + 2x + 4x^2/2 + 8x^3/6 + 16x^4/24 + ... e^(2x) = 1 + 2x + 2x^2 + 4x^3/3 + 2x^4/3 + ... Step 2: Compute the numerator e^(2x) - 1 - 2x. e^(2x) - 1 - 2x = (1 + 2x + 2x^2 + 4x^3/3 + ...) - 1 - 2x e^(2x) - 1 - 2x = 2x^2 + 4x^3/3 + O(x^4) Step 3: Expand sin x using its Taylor series. sin x = x - x^3/3! + x^5/5! - ... sin x = x - x^3/6 + x^5/120 - ... Step 4: Compute x^2 sin x. x^2 sin x = x^2(x - x^3/6 + x^5/120 - ...) x^2 sin x = x^3 - x^5/6 + x^7/120 - ... Step 5: Form the ratio and simplify. [(e^(2x) - 1 - 2x) / (x^2 sin x)] = [2x^2 + 4x^3/3 + O(x^4)] / [x^3 - x^5/6 + ...] Factor out x^2 from the numerator and x^3 from the denominator: = [x^2(2 + 4x/3 + O(x^2))] / [x^3(1 - x^2/6 + ...)] = [2 + 4x/3 + O(x^2)] / [x(1 - x^2/6 + ...)] Step 6: Analyze the limit. As x → 0, the numerator approaches 2, while the denominator is x times a factor approaching 1. This means the denominator approaches 0 while the numerator approaches 2. Let me reconsider: I need to be more careful with the order of terms. Actually, from Step 2: e^(2x) - 1 - 2x = 2x^2 + 4x^3/3 + O(x^4) From Step 4: x^2 sin x = x^3 - x^5/6 + O(x^7) So the ratio is: [2x^2 + 4x^3/3 + O(x^4)] / [x^3 - x^5/6 + O(x^7)] = [x^2(2 + 4x/3 + O(x^2))] / [x^3(1 - x^2/6 + O(x^4))] = (1/x) · [2 + 4x/3 + O(x^2)] / [1 - x^2/6 + O(x^4)] This suggests the limit diverges. Let me verify the numerator expansion once more. e^(2x) = 1 + 2x + 2x^2 + (4/3)x^3 + (2/3)x^4 + ... e^(2x) - 1 - 2x = 2x^2 + (4/3)x^3 + (2/3)x^4 + ... This is correct. The numerator starts at x^2 while the denominator x^2 sin x starts at x^3, so the limit is indeed infinite. Wait—let me recalculate x^2 sin x more carefully: x^2 sin x = x^2 · (x - x^3/6 + ...) = x^3 - x^5/6 + ... So we have: [2x^2 + (4/3)x^3 + ...] / [x^3 - x^5/6 + ...] Dividing numerator and denominator by x^3: [2/x + 4/3 + ...] / [1 - x^2/6 + ...] As x → 0+, this diverges to +∞, and as x → 0-, this diverges to -∞. However, if the problem intends a finite answer, let me check if there's an error. Upon reflection, the limit does not exist in the finite sense; it diverges.