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Question Number 14152 by tawa tawa last updated on 28/May/17

Answered by ajfour last updated on 29/May/17

$$L=\underset{\theta \to \pi /6}{\lim }\left[\frac{\tan {}^2\theta +\tan {}^4\theta +\tan {}^6\theta +...}{(1+\tan {}^2\theta -\frac{5}{6})+{(1+\tan {}^2\theta -\frac{5}{6})}^2+...}\right]$$$$L=\frac{\frac{1}{3}+{\left(\frac{1}{3}\right)}^2+{\left(\frac{1}{3}\right)}^3+...}{\frac{1}{2}+{\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^3+...}$$$$L=\frac{\frac{1}{3}\left(\frac{1}{1-1/3}\right)}{\frac{1}{2}\left(\frac{1}{1-1/2}\right)}=\frac{\frac{1}{3}\times \frac{3}{2}}{\frac{1}{2}\times \frac{2}{1}}=\frac{1}{2}.$$$$$$

Commented by tawa tawa last updated on 29/May/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{I}\mathrm{really}\mathrm{appreciate}.$$

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