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Limits: Question

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Question
lim⁡x→0 ⁣(sin⁡xx) ⁣ ⁣1/x2\displaystyle\lim_{x\to 0}\!\left(\dfrac{\sin x}{x}\right)^{\!\!1/x^{2}}
Ae1/2e^{1/2}
B11
Ce−1/6e^{-1/6}correct
De1/3e^{1/3}
Solution
**Step 1: Identify the form.** sin⁡xx→1\dfrac{\sin x}{x}\to 1 and 1x2→∞\dfrac{1}{x^{2}}\to\infty. Form is 1∞1^{\infty}. **Step 2: Standard formula for 1∞1^{\infty}.**
lim⁡f(x)g(x)=exp⁡ ⁣[lim⁡g(x) ⁣(f(x)−1)]\lim f(x)^{g(x)}=\exp\!\left[\lim g(x)\!\left(f(x)-1\right)\right]
 **Step 3: Apply with f(x)=sin⁡xx,  g(x)=1x2f(x)=\dfrac{\sin x}{x},\;g(x)=\dfrac{1}{x^{2}}.**
L=exp⁡ ⁣[lim⁡x→01x2 ⁣(sin⁡xx−1)]=exp⁡ ⁣[lim⁡x→0sin⁡x−xx3]L=\exp\!\left[\lim_{x\to 0}\dfrac{1}{x^{2}}\!\left(\dfrac{\sin x}{x}-1\right)\right]=\exp\!\left[\lim_{x\to 0}\dfrac{\sin x-x}{x^{3}}\right]
 **Step 4: Use Taylor expansion sin⁡x=x−x36+…\sin x=x-\dfrac{x^{3}}{6}+\ldots.**
sin⁡x−xx3=−x3/6+…x3→−16\dfrac{\sin x-x}{x^{3}}=\dfrac{-x^{3}/6+\ldots}{x^{3}}\to -\dfrac{1}{6}
 **Step 5: Substitute.**
L=e−1/6L=e^{-1/6}
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