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Question Number 22165 by A1B1C1D1 last updated on 12/Oct/17

Answered by ajfour last updated on 13/Oct/17

$$let\frac{x}{1-x^2}=t\Rightarrow x^2=1-\frac{x}{t}$$$$L=\underset{x\to 0}{\lim }\left(\frac{1+\frac{x}{t}-2(1-\frac{t^2}{2}+\frac{t^4}{24}-....)}{2x^4}\right)$$$$=\underset{x\to 0}{\lim }\left(\frac{-1+\frac{x}{t}+t^2-\frac{t^4}{12}+....}{2x^4}\right)$$$$=\underset{x\to 0}{\lim }\left(\frac{-x^2+t^2(1-\frac{t^2}{12}+....)}{2x^4}\right)$$$$=\underset{x\to 0}{\lim }\left(\frac{-x^2+\frac{x^2}{{(1-x^2)}^2}(1-\frac{t^2}{12}+...)}{2x^4}\right)$$$$=\underset{x\to 0}{\lim }\left(\frac{-1+\frac{1}{{(1-x^2)}^2}(1-\frac{t^2}{12}+...)}{2x^2}\right)$$$$=\underset{x\to 0}{\lim }\left(\frac{-1+{(1-x^2)}^{-2}(1-\frac{t^2}{12}+..)}{2x^2}\right)$$$$=\underset{x\to 0}{\lim }\left(\frac{-1+(1+2x^2+..)(1-\frac{t^2}{12}+..)}{2x^2}\right)$$$$=\underset{x\to 0}{\lim }\left(\frac{-1+1+2x^2-\frac{t^2}{12}+...}{2x^2}\right)$$$$=\underset{x\to 0}{\lim }\left(\frac{2x^2-\frac{x^2}{12{(1-x^2)}^2}+...}{2x^2}\right)$$$$=\underset{x\to 0}{\lim }\left(\frac{2-\frac{1}{12{(1-x^2)}^2}+...}{2}\right)$$$$L=\frac{2-\frac{1}{12}}{2}=\frac{23}{24}.$$$$$$

Commented by A1B1C1D1 last updated on 13/Oct/17

$$\mathrm{Thank}\mathrm{you}.$$

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