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

 Jun 6, 2018

Best Answer 

 #1
avatar+971 
+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

 Jun 6, 2018
 #1
avatar+971 
+2
Best Answer

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

 

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

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