evaluate the following limit. lim x→0+ (sin x / x)^(12 / x^2) lim x→0+ (sin x / x)^(12 / x^2) =□ (type an…

Explanation:

Step1: Recall the well - known limit

We know that $\lim_{x\rightarrow0}\frac{\sin x}{x}=1$. We also use the fact that if $y = (\frac{\sin x}{x})^{\frac{12}{x^{2}}}$, then $\ln y=\frac{12}{x^{2}}\ln(\frac{\sin x}{x})$.

Step2: Use the Taylor series expansion

The Taylor series of $\sin x=x-\frac{x^{3}}{6}+\frac{x^{5}}{120}-\cdots$. So, $\frac{\sin x}{x}=1 - \frac{x^{2}}{6}+\frac{x^{4}}{120}-\cdots$. Then, $\ln(\frac{\sin x}{x})=\ln(1 - \frac{x^{2}}{6}+\frac{x^{4}}{120}-\cdots)$. For small $u$, $\ln(1 + u)\approx u$ when $|u|\ll1$. Let $u=-\frac{x^{2}}{6}+\frac{x^{4}}{120}-\cdots$. Then $\ln(\frac{\sin x}{x})\approx-\frac{x^{2}}{6}$ as $x\rightarrow0$.

Step3: Find the limit of $\ln y$

$\lim_{x\rightarrow0^{+}}\ln y=\lim_{x\rightarrow0^{+}}\frac{12}{x^{2}}\ln(\frac{\sin x}{x})$. Substituting $\ln(\frac{\sin x}{x})\approx-\frac{x^{2}}{6}$ into the limit, we get $\lim_{x\rightarrow0^{+}}\frac{12}{x^{2}}\times(-\frac{x^{2}}{6})=- 2$.

Step4: Find the limit of $y$

Since $\lim_{x\rightarrow0^{+}}\ln y=-2$, and $y = e^{\ln y}$, then $\lim_{x\rightarrow0^{+}}y = e^{-2}=\frac{1}{e^{2}}$.

Answer:

$\frac{1}{e^{2}}$