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Question Number 216638 by Nadirhashim last updated on 13/Feb/25

$$\mathit{w}\mathit{i}\mathit{t}\mathit{h}\mathit{o}\mathit{u}\mathit{t}\mathit{u}\mathit{s}\mathit{i}\mathit{n}\mathit{g}\mathit{L}\mathit{H}\mathit{o}\mathit{p}\mathit{i}\mathit{t}\mathit{a}\mathit{l}$$$$\mathit{r}\mathit{u}\mathit{l}\mathit{e}\mathit{e}\mathit{v}\mathit{a}\mathit{l}\mathit{u}\mathit{t}\mathit{e}$$$$\underset{x\to 0}{\lim }\frac{\mathit{l}\mathit{n}(1-x)-\mathit{s}\mathit{i}\mathit{n}\left(\mathit{x}\right)}{1-{\mathit{c}\mathit{o}\mathit{x}}^2\left(\mathit{x}\right)}$$

Commented by MathematicalUser2357 last updated on 13/Feb/25

$$typoindenominator$$

Answered by mahdipoor last updated on 13/Feb/25

$$\frac{(-x-\frac{x^2}{2}-\frac{x^3}{3}-...)-(x-\frac{x^3}{3!}+\frac{x^5}{5!}-...)}{1-{(1-\frac{x^2}{2!}+\frac{x^4}{4}-..)}^2}$$$$=\frac{x(a+bx+...)}{x^2(A+Bx+..)}\Rightarrow limx\to 0=\pm \mathrm{\infty }$$

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