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Question Number 125649 by bramlexs22 last updated on 12/Dec/20

$$\underset{x\to 0}{\lim }\frac{\ln (1+x\mathrm{arc}\tan x)+1-e^x}{\sqrt{1+2x^4}-1}?$$

Answered by Olaf last updated on 13/Dec/20

$$\mathrm{Let}f\left(x\right)=\frac{\ln (1+x\arctan x)+1-e^x}{\sqrt{1+2x^4}-1}$$$$f\left(x\right)\underset{0}{\sim }\frac{\ln (1+x(x-\frac{x^3}{3}))+1-e^x}{\sqrt{1+2x^4}-1}$$$$f\left(x\right)\underset{0}{\sim }\frac{(x^2-\frac{x^4}{3})+1-(1+x+\frac{x^2}{2})}{(1+x^4)-1}$$$$f\left(x\right)\underset{0}{\sim }\frac{\frac{1}{2}{x}^2-x}{x^4}\to \mathrm{\infty }$$$$$$

Commented by bramlexs22 last updated on 13/Dec/20

$$theanswer-\frac{4}{3}sir$$

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