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Question Number 114543 by bemath last updated on 19/Sep/20

$$\underset{x\to 0}{\lim }\frac{x^2\arctan \left(2x\right)}{x\cos x+\tan x}?$$

Answered by bobhans last updated on 19/Sep/20

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

Answered by mathmax by abdo last updated on 19/Sep/20

$$\mathrm{by}\mathrm{hospital}\mathrm{theorem}\mathrm{we}\mathrm{get}$$$$\mathrm{L}={\lim }_{\mathrm{x}\to 0}\frac{2\mathrm{x}\mathrm{arctanx}+{\mathrm{x}}^2.\frac{2}{1+{4\mathrm{x}}^2}}{\mathrm{cosx}-\mathrm{xsinx}+1+{\tan }^2\mathrm{x}}=\frac{0}{2}=0$$

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