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Question Number 203651 by Davidtim last updated on 24/Jan/24

$$\underset{x\to 1}{\lim }{(2-x)}^{tan\left(\frac{\pi x}{2}\right)}=?$$

Answered by Mathspace last updated on 24/Jan/24

$$changement1-x=tgive$$$$f\left(x\right)={(2-x)}^{tan\left(\frac{\pi x}{2}\right)}$$$$=f(1-t)={(2-(1-t))}^{tan\left(\frac{\pi }{2}(1-t)\right)}$$$$={(1+t)}^{tan(\frac{\pi }{2}-\frac{\pi t}{2})}$$$$={(1+t)}^{\frac{1}{tan\left(\frac{\pi t}{2}\right)}}(x\to 1\iff t\to 0)$$$$ln\left(f(1-t)\right)=\frac{1}{tan\left(\frac{\pi t}{2}\right)}ln(1+t)$$$$\sim \frac{2}{\pi t}\times t(t\in V\left(0\right)$$$$=\frac{2}{\pi }so$$$$lim(ln\left(f(1-t)\right)=\frac{2}{\pi }\Rightarrow$$$${lim}_{t\to 0}f(1-t)=e^{\frac{2}{\pi }}$$$$\Rightarrow {lim}_{x\to 1}f\left(x\right)=e^{\frac{2}{\pi }}$$

Answered by MM42 last updated on 24/Jan/24

$$1-x=u$$$$\Rightarrow {lim}_{u\to 0}{(1+u)}^{cot\left(\frac{\pi }{2}u\right)}$$$$\Rightarrow {lim}_{u\to 0}{\left({(1+u)}^{\frac{1}{u}}\right)}^{\frac{u}{sin\frac{\pi u}{2}}\times cos\frac{\pi u}{2}}$$$$=e^{\frac{2}{\pi }}✓$$

Answered by Calculusboy last updated on 26/Jan/24

$$\mathit{S}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}:\mathit{b}\mathit{y}\mathit{s}\mathit{u}\mathit{b}\mathit{d}\mathit{i}\mathit{r}\mathit{e}\mathit{c}\mathit{t}\mathit{l}\mathit{y},\mathit{w}\mathit{e}\mathit{g}\mathit{e}\mathit{t}\left(1^{\mathrm{\infty }}\right)\mathit{i}\mathit{n}\mathit{d}\mathit{e}\mathit{t}\mathit{e}\mathit{r}\mathit{m}\mathit{i}\mathit{n}\mathit{a}\mathit{n}\mathit{t}$$$$\mathit{l}\mathit{e}\mathit{t}\mathit{y}={(2-\mathit{x})}^{\mathit{t}\mathit{a}\mathit{n}\left(\frac{\mathit{\pi }\mathit{x}}{2}\right)}\left(\mathit{a}\mathit{p}\mathit{p}\mathit{l}\mathit{y}\mathit{l}\mathit{o}\mathit{g}\mathit{t}\mathit{o}\mathit{b}\mathit{o}\mathit{t}\mathit{h}\mathit{s}\mathit{i}\mathit{d}\mathit{e}\mathit{s}\right)$$$$\mathit{l}\mathit{o}\mathit{g}\mathit{y}=\mathit{l}\mathit{o}\mathit{g}{(2-\mathit{x})}^{\mathit{t}\mathit{a}\mathit{n}\left(\frac{\mathit{\pi }\mathit{x}}{2}\right)}$$$$\mathit{l}\mathit{o}\mathit{g}\mathit{y}=\mathit{l}\mathit{i}\underset{\mathit{x}\to 1}{\mathit{m}\mathit{t}\mathit{a}\mathit{n}}\left(\frac{\mathit{\pi }\mathit{x}}{2}\right)\mathit{l}\mathit{o}\mathit{g}(2-\mathit{x})$$$$\mathit{l}\mathit{o}\mathit{g}\mathit{y}=\mathit{l}\mathit{i}\underset{\mathit{x}\to 1}{\mathit{m}}\frac{\frac{\mathit{d}}{\mathit{d}\mathit{x}}\left[\mathit{l}\mathit{o}\mathit{g}(2-\mathit{x})\right]}{\frac{\mathit{d}}{\mathit{d}\mathit{x}}\left[\frac{1}{\mathit{t}\mathit{a}\mathit{n}\left(\frac{\mathit{\pi }\mathit{x}}{2}\right)}\right]}\Rightarrow \mathit{l}\mathit{o}\mathit{g}\mathit{y}=\mathit{l}\mathit{i}\underset{\mathit{x}\to 1}{\mathit{m}}\frac{\frac{-1}{2-\mathit{x}}}{\frac{\mathit{d}}{\mathit{d}\mathit{x}}\left[\mathit{c}\mathit{o}\mathit{t}\left(\frac{\mathit{\pi }\mathit{x}}{2}\right)\right]}$$$$\mathit{l}\mathit{o}\mathit{g}\mathit{y}=\mathit{l}\mathit{i}\underset{\mathit{x}\to 1}{\mathit{m}}\frac{\frac{-1}{2-\mathit{x}}}{[-\frac{\mathit{\pi }}{2}{\mathit{c}\mathit{o}\mathit{s}\mathit{e}\mathit{c}}^2\left(\frac{\mathit{\pi }\mathit{x}}{2}\right)]}$$$$\mathit{l}\mathit{o}\mathit{g}\mathit{y}=\frac{\frac{-1}{1}}{-\frac{\mathit{\pi }}{2}\mathit{l}\mathit{i}\underset{\mathit{x}\to 1}{\mathit{m}}{\left[\mathit{c}\mathit{o}\mathit{s}\mathit{e}\mathit{c}\left(\frac{\mathit{\pi }\mathit{x}}{2}\right)\right]}^2}$$$$\mathit{l}\mathit{o}\mathit{g}\mathit{y}=\frac{-1}{-\frac{\mathit{\pi }}{2}\cdot 1}\Rightarrow \mathit{l}\mathit{o}\mathit{g}\mathit{y}=\frac{2}{\mathit{\pi }}$$$$\mathit{y}={\mathit{e}}^{\frac{2}{\mathit{\pi }}}$$$$\therefore \mathit{l}\mathit{i}\underset{\mathit{x}\to 1}{\mathit{m}}{(2-\mathit{x})}^{\mathit{t}\mathit{a}\mathit{n}\left(\frac{\mathit{\pi }\mathit{x}}{2}\right)}={\mathit{e}}^{\frac{2}{\mathit{\pi }}}$$$$$$

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