Triogo-limit

**dhiab** · June 1st 2010, 10:19 PM

Calculate : $\150dpi \lim_{x\rightarrow 1}\frac{sin\left ( \frac{\pi }{2x} \right )-1}{x^{3}-1}$

**Prove It** · June 1st 2010, 10:25 PM

Originally Posted by dhiab

Calculate : $\150dpi \lim_{x\rightarrow 1}\frac{sin\left ( \frac{\pi }{2x} \right )-1}{x^{3}-1}$

Since this $\to \frac{0}{0}$ you can use L'Hospital's Rule...

$\lim_{x \to 1}\frac{\sin{\left(\frac{\pi}{2x}\right)} - 1}{x^3 - 1} = \lim_{x \to 1}\frac{-\frac{\pi}{2x^2}\cos{\left(\frac{\pi}{2x}\right)}} {3x^2}$

$= \frac{-\frac{\pi}{2(1)^2}\cos{\left[\frac{\pi}{2(1)}\right]}}{3(1)^2}$

$= 0$ .

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