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Question Number 214256 by Einstein2006 last updated on 02/Dec/24

$${\mathit{l}\mathit{i}\mathit{m}}_{\mathit{x}\to 1}\propto .\mathit{a}\mathit{r}\mathit{c}\mathit{t}\mathit{a}\mathit{n}(\frac{2}{1+\mathit{x}}-1)$$$$\bullet \mathit{C}\mathit{a}\mathit{l}\mathit{c}\mathit{u}\mathit{l}\mathit{o}\mathit{n}\mathit{s}\mathit{l}\mathit{a}\mathit{l}\mathit{i}\mathit{m}\mathit{i}\mathit{t}\mathit{e}\mathit{a}{\mathit{l}}^{\prime }\mathit{i}\mathit{n}\mathit{t}\mathit{r}\mathit{i}\mathit{e}\mathit{u}\mathit{r}:$$$$\frac{2}{1+\mathit{x}}-1=0$$$$$$$${\mathit{l}\mathit{i}\mathit{m}}_{\mathit{x}\to 1}arctan(\frac{2}{1+x}-1)=\mathit{a}\mathit{r}\mathit{c}\mathit{t}\mathit{a}\mathit{n}\left(0\right)=0$$$$$$$${\mathit{l}\mathit{i}\mathit{m}}_{\mathit{x}\to 1}\propto .arctan(\frac{2}{1+x}-1)=\propto .0$$$$$$$${\mathit{l}\mathit{i}\mathit{m}}_{\mathit{x}\to 1}\propto .\mathit{a}\mathit{r}\mathit{c}\mathit{t}\mathit{a}\mathit{n}(\frac{2}{1+\mathit{x}}-1)=0$$

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