# Find the limit of: <munder> <mo movablelimits="true" form="prefix">lim <mrow class="

Find the limit of:
$\underset{x\to \frac{\pi }{3}}{lim}\frac{1-2\mathrm{cos}x}{\pi -3x}$
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AlokMoopisppf
Use l'Hôpital's rule:
$\underset{x\to \frac{\pi }{3}}{lim}\frac{1-2\mathrm{cos}\left(x\right)}{\pi -3x}=\underset{x\to \frac{\pi }{3}}{lim}\frac{2\mathrm{sin}\left(x\right)}{-3}=-\frac{2}{3}\underset{x\to \frac{\pi }{3}}{lim}\mathrm{sin}\left(x\right)=-\frac{2}{3}\cdot \mathrm{sin}\left(\frac{\pi }{3}\right)=-\frac{2}{3}\cdot \frac{\sqrt{3}}{2}=-\frac{1}{\sqrt{3}}$
###### Not exactly what you’re looking for?
tomekmusicd9
Put $x=\frac{\pi }{3}+h$. Then the given limit is
$\begin{array}{rl}\underset{h\to 0}{lim}\frac{1-2\mathrm{cos}\left(\frac{\pi }{3}+h\right)}{-3h}& =\underset{h\to 0}{lim}\frac{1-\mathrm{cos}h+\sqrt{3}\mathrm{sin}h}{-3h}\\ & =\underset{h\to 0}{lim}\frac{2{\mathrm{sin}}^{2}\frac{h}{2}}{-3h}+\underset{h\to 0}{lim}\sqrt{3}\frac{\mathrm{sin}h}{-3h}\\ & =-\frac{1}{\sqrt{3}}\end{array}$