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Question Number 99720 by bobhans last updated on 23/Jun/20

$$\underset{x\to 0}{\lim }\frac{1}{{\mathrm{x}}^2}-\frac{1}{\tan {}^2\mathrm{x}}?$$

Commented by john santu last updated on 23/Jun/20

$$\underset{x\to 0}{\lim }\frac{\tan {}^2\mathrm{x}-{\mathrm{x}}^2}{{\mathrm{x}}^2\tan {}^2\mathrm{x}}=$$$$\underset{x\to 0}{\lim }\frac{(\tan \mathrm{x}+\mathrm{x})(\tan \mathrm{x}-\mathrm{x})}{{\mathrm{x}}^2\tan {}^2\mathrm{x}}=$$$$\underset{x\to 0}{\lim }\frac{(\mathrm{x}+\frac{{\mathrm{x}}^3}{3}+\mathrm{x})(\mathrm{x}+\frac{{\mathrm{x}}^3}{3}-\mathrm{x})}{{\mathrm{x}}^4}=$$$$\underset{x\to 0}{\lim }\frac{\mathrm{x}(2+\frac{{\mathrm{x}}^2}{3})\left(\frac{{\mathrm{x}}^3}{3}\right)}{{\mathrm{x}}^4}=\frac{2}{3}$$

Commented by mr W last updated on 23/Jun/20

$$=\frac{1}{x^2}-(\frac{1}{x^2}-\frac{2}{3}+\frac{x^2}{15}-...)$$$$\to \frac{2}{3}$$

Commented by john santu last updated on 23/Jun/20

$$\mathrm{joos}...\mathrm{tooss}\mathrm{prof}\mathrm{W}$$

Commented by bobhans last updated on 24/Jun/20

$$\mathrm{sir}\mathrm{w},\mathrm{how}\mathrm{you}\mathrm{get}\frac{1}{{\mathrm{x}}^2}-\frac{2}{3}+\frac{{\mathrm{x}}^2}{15}-...?$$

Commented by mr W last updated on 24/Jun/20

$$Maclaurinseriesof\frac{1}{{\tan }^{2}x}$$

Answered by Ar Brandon last updated on 23/Jun/20

$$l=\underset{\mathrm{x}\to 0}{\lim }\{\frac{1}{{\mathrm{x}}^2}-\frac{1}{{\tan }^2\mathrm{x}}\}=\underset{\mathrm{x}\to 0}{\lim }\{\frac{1}{{\mathrm{x}}^2}-\frac{{\cos }^2\mathrm{x}}{{\sin }^2\mathrm{x}}\}=\underset{\mathrm{x}\to 0}{\lim }\left\{\frac{{\sin }^2\mathrm{x}-{\mathrm{x}}^2{\cos }^2\mathrm{x}}{{\mathrm{x}}^2{\sin }^2\mathrm{x}}\right\}$$$$=\underset{\mathrm{x}\to 0}{\lim }\left\{\frac{2\mathrm{sinxcosx}-(-{2\mathrm{x}}^2\mathrm{sinxcosx}+{2\mathrm{xcos}}^2\mathrm{x})}{{2\mathrm{x}}^2\mathrm{sinxcosx}+{2\mathrm{xsin}}^2\mathrm{x}}\right\}$$$$Itriedthisbutobviouslyit{isn}^{\prime }tthebestapproach.$$$$Youmayliketotry{Maclaurine}^{\prime }sexpansion.$$

Commented by bobhans last updated on 23/Jun/20

$$\mathrm{yes}.{\mathrm{L}}^{\prime }\mathrm{hopital}\mathrm{maybe}\mathrm{alternative}\mathrm{method}$$

Answered by john santu last updated on 23/Jun/20

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