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Question Number 176213 by cortano1 last updated on 15/Sep/22

$$\underset{x\to 0}{\lim }\frac{\arctan \mathrm{x}}{\arcsin \mathrm{x}-\mathrm{x}}=?$$

Answered by flamable last updated on 15/Sep/22

$$$$$$BythehelpofTaylor-Youngformulaeat$$$$x=0wehavethefollowing:$$$$\frac{1}{1+x}=1-x+x^2-x^3+...+{(-1)}^n{x}^{n}n\in \mathbb{N}$$$$whenx=t^2$$$$\frac{1}{1+t^2}=1-t^2+t^4-t^6+...$$$$\Rightarrow arctan\left(t\right)=t-\frac{1}{3}{t}^3+t^3\epsilon \left(t\right),\epsilon \left(t\right)\to 0_{t\to 0}$$$$\frac{1}{1-x}=1+x+x^2+x^3+...+x^{n}n\in \mathbb{N}$$$$whenx=t^2$$$$\frac{1}{1-t^2}=1+t^2+t^4+...+t^{2n}$$$$\frac{1}{\sqrt{1-t^2}}=1-\frac{1}{2}{t}^2+\frac{3}{8}{t}^4+...$$$$\Rightarrow arcsin\left(t\right)=t-\frac{1}{6}{t}^3+t^3\epsilon \left(t\right),\epsilon \left(t\right)\to \underset{t\to 0}{0}$$$$\underset{t\to 0}{\lim }\frac{arctan\left(t\right)}{arcsin\left(t\right)-t}=\underset{t\to 0}{\lim }\frac{t-\frac{1}{3}{t}^3+t^3\epsilon \left(t\right)}{t-\frac{1}{6}{t}^3+t^3\epsilon \left(t\right)-t}$$$$=\underset{t\to 0}{\lim }-\frac{t(1-\frac{1}{3}{t}^2+t^2\epsilon \left(t\right))}{t(\frac{1}{6}{t}^2+t^2\epsilon \left(t\right))}$$$$=-1$$$$\therefore \underset{t\to 0}{\lim }\frac{arctan\left(t\right)}{arcsin\left(t\right)-t}=-1$$$$$$$$$$

Commented by cortano1 last updated on 16/Sep/22

$$\mathrm{No}$$

Commented by Tawa11 last updated on 25/Sep/22

$$\mathrm{Great}\mathrm{sir}$$

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