How to compute $\displaystyle \lim_{x\to\frac{2}{3}}\frac{x^2 \cos(\pi/x)}{x - \frac{2}{3}}$? I'm stuck in this question. Is there a way not using the l'Hôpital's rule?
How to compute $\lim_{x\to\frac{2}{3}}\frac{x^2 \cos(\pi/x)}{x - \frac{2}{3}}$
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calculus
limits
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2The shortest way is to use the Taylor expansion of $f(x) = \cos (\dfrac{\pi}{x})$ at $x = \dfrac{2}{3}$ up to second term and you should be done. – 2017-01-31
3 Answers
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We have $$ \lim_{x\to\frac{2}{3}}\frac{x^2 \cos( \frac{\pi}{x})}{x - \frac{2}{3}} = \lim_{x\to\frac{2}{3}}\frac{x^2 \cos(\frac{\pi}{x}) - \left(\frac{2}{3} \right)^2 \cos \left( \frac{3\pi}{2}\right)}{x - \frac{2}{3}} = \left.\frac{\mathrm d}{\mathrm d x}x^2 \cos \left(\frac{\pi}{x} \right) \right|_{x=\frac{2}{3}} = -\pi.$$
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From $\cos\alpha=-\sin(\dfrac{3\pi}{2}-\alpha)$ write \begin{eqnarray} \lim_{x\to\frac{2}{3}}\frac{x^2 \cos(\dfrac{\pi}{x})}{x - \dfrac{2}{3}} &=& \lim_{x\to\frac{2}{3}}\frac{x^2 \cos(\dfrac{\pi}{x})}{x - \dfrac{2}{3}}\\ &=& \lim_{x\to\frac{2}{3}}\frac{-x\sin(\dfrac{3\pi}{2}-\dfrac{\pi}{x})}{1-\dfrac{2}{3x}}\\ &=& \lim_{x\to\frac{2}{3}}\frac{-x\sin(\pi(\dfrac{3}{2}-\dfrac{1}{x}))}{\dfrac{2}{3}(\dfrac{3}{2}-\dfrac{1}{x})}\\ &=& \lim_{x\to\frac{2}{3}}\frac{-x\pi(\dfrac{3}{2}-\dfrac{1}{x})}{\dfrac{2}{3}(\dfrac{3}{2}-\dfrac{1}{x})}\\ &=& \color{blue}{-\pi} \end{eqnarray}
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Set $x-2/3=y$
$\cos \pi/x=\cos 3\pi/(3y+2)=-\sin(3\pi/2- 3\pi/(3y+2))$