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# ​ Compute, calculus help

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$$\displaystyle\lim_{x\to0}(\frac{\sin{x}}{x})^\frac{1}{1-\cos{x}}$$

Jun 6, 2018

#1
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$$\text{Compute}:\displaystyle\lim_{x\to0}(\frac{\sin{x}}{x})^\frac{1}{1-\cos{x}}$$

We can approximate $$\sin{x} \text{ and } \cos{x}$$ by their Taylor series.

After apply substitutions:

$$\displaystyle\lim_{x\to0}(\frac{\sin{x}}{x})^\frac{1}{1-\cos{x}}=\lim_{x\to0}(1-\frac{x^2}{6})^\frac{2}{x^2}=\lim_{x\to0}(1+\frac{x^2}{6})^{-\frac{2}{x^2}}$$

$$=\displaystyle\lim_{x\to\infty}(1+\frac{1}{6x^2})^{-2x^2}=\lim_{x\to\infty}(1+\frac{1}{x})^{-x/3}=\boxed{e^{-1/3}}$$

This is because $$e=\displaystyle\lim_{x\to\infty}(1+\frac1x)^x$$

I hope this helped,

Gavin

Jun 6, 2018

#1
+986
+2

$$\text{Compute}:\displaystyle\lim_{x\to0}(\frac{\sin{x}}{x})^\frac{1}{1-\cos{x}}$$

We can approximate $$\sin{x} \text{ and } \cos{x}$$ by their Taylor series.

After apply substitutions:

$$\displaystyle\lim_{x\to0}(\frac{\sin{x}}{x})^\frac{1}{1-\cos{x}}=\lim_{x\to0}(1-\frac{x^2}{6})^\frac{2}{x^2}=\lim_{x\to0}(1+\frac{x^2}{6})^{-\frac{2}{x^2}}$$

$$=\displaystyle\lim_{x\to\infty}(1+\frac{1}{6x^2})^{-2x^2}=\lim_{x\to\infty}(1+\frac{1}{x})^{-x/3}=\boxed{e^{-1/3}}$$

This is because $$e=\displaystyle\lim_{x\to\infty}(1+\frac1x)^x$$

I hope this helped,

Gavin

GYanggg Jun 6, 2018