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Question Number 115777 by bemath last updated on 28/Sep/20

$$\underset{x\to 0}{\lim }\frac{\sin x-{\tan }^{-1}x}{x^2\ln (1+x)}$$

Answered by bobhans last updated on 28/Sep/20

$$\underset{x\to 0}{\lim }\frac{(x-\frac{x^3}{6}+{Px}^5)-(x-\frac{x^3}{3}+{Qx}^5)}{x^2(x-\frac{1}{2}{x}^2+{Rx}^3)}=$$$$\underset{x\to 0}{\lim }\frac{\frac{1}{6}{x}^3+(P-Q)x^5}{x^3-\frac{1}{2}{x}^4+{Rx}^5}=$$$$\underset{x\to 0}{\lim }\frac{\frac{1}{6}+(P-Q)x^2}{1-\frac{1}{2}x+{Rx}^2}=\frac{1}{6}$$

Answered by Olaf last updated on 28/Sep/20

$$=\underset{x\to 0}{\lim }\frac{(x-\frac{x^3}{6})-(x-\frac{x^3}{3})}{x^2(x-\frac{x^2}{2})}$$$$=\underset{x\to 0}{\lim }\frac{-\frac{x^3}{6}+\frac{x^3}{3}}{x^2(x-\frac{x^2}{2})}=-\frac{1}{6}+\frac{1}{3}=\frac{1}{6}$$$$$$

Answered by Dwaipayan Shikari last updated on 28/Sep/20

$$\underset{x\to 0}{\lim }\frac{(\mathrm{x}-\frac{{\mathrm{x}}^3}{6})-(\mathrm{x}-\frac{{\mathrm{x}}^3}{3})}{{\mathrm{x}}^2.\mathrm{x}}=\frac{\frac{{\mathrm{x}}^3}{6}}{{\mathrm{x}}^3}=\frac{1}{6}\underset{x\to 0}{\lim }\log (1+\mathrm{x})=\mathrm{x}$$

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