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Question Number 66321 by mathmax by abdo last updated on 12/Aug/19

$$find{lim}_{x\to 0^{+}}{\left(tan\left(\frac{\pi }{2+x}\right)\right)}^x$$

Commented by mathmax by abdo last updated on 24/Aug/19

$$letA\left(x\right)={\left(tan\left(\frac{\pi }{2+x}\right)\right)}^x\Rightarrow A\left(x\right)=e^{xln\left(\frac{\pi }{2+x}\right)}$$$$changement\frac{\pi }{2+x}=tgive\frac{2+x}{\pi }=\frac{1}{t}\Rightarrow 2+x=\frac{\pi }{t}\Rightarrow x=\frac{\pi }{t}-2\Rightarrow$$$$A\left(x\right)=B\left(t\right)=e^{(\frac{\pi }{t}-2)ln\left(tant\right)}\Rightarrow {lim}_{x\to 0}A\left(x\right)={lim}_{t\to \frac{\pi }{2}}{e}^{(\frac{\pi }{t}-2)ln\left(t\right)}$$$$={lim}_{t\to \frac{\pi }{2}}{e}^{(\pi -2t)\frac{ln\left(tant\right)}{2t}}changement\pi -2t=ugive2t=\pi -u$$$${lim}_{t\to \frac{\pi }{2}}B\left(t\right)={lim}_{u\to 0}{e}^{u\frac{ln\left(tan(\frac{\pi }{2}-\frac{u}{2})\right)}{\pi -u}}$$$$={lim}_{u\to 0}{e}^{\frac{u}{\pi -u}ln\left(\frac{1}{tan\left(\frac{u}{2}\right)}\right)}={lim}_{u\to 0}{e}^{-\frac{u}{\pi -u}ln\left(tan\left(\frac{u}{2}\right)\right)}=1$$$$becausetan\left(\frac{u}{2}\right)\sim \frac{u}{2}\left(V\left(0\right)\right)and{lim}_{u\to 0^{+}}ulnu=0$$

Commented by mathmax by abdo last updated on 24/Aug/19

$$anotherwaywehave\frac{\pi }{2+x}=\frac{\pi }{2(1+\frac{x}{2})}\sim \frac{\pi }{2}(1-\frac{x}{2})=\frac{\pi }{2}-\frac{\pi x}{4}$$$$\Rightarrow tan\left(\frac{\pi }{2+x}\right)=\frac{1}{tan\left(\frac{\pi x}{4}\right)}but{\left(tan\left(\frac{\pi }{2+x}\right)\right)}^x=e^{xln\left(tan\left(\frac{\pi }{2+x}\right)\right)}$$$$\Rightarrow f\left(x\right)=e^{-xln\left(tan\left(\frac{\pi x}{4}\right)\right)}\sim e^{-xln\left(\frac{\pi x}{4}\right)}\to 1(x\to 0)$$

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