Brock Brown

2022-01-02

How do I evaluate $\underset{z\to 1}{lim}\left(1-z\right)\mathrm{tan}\frac{\pi z}{2}$
I tried using the identity, $\mathrm{tan}\frac{x}{2}=\frac{1-\mathrm{cos}x}{\mathrm{sin}x}$ to simplify this to:
$\underset{z\to 1}{lim}\left(1-z\right)\frac{\mathrm{sin}\pi z}{1+\mathrm{cos}\pi z}$

psor32

Note that
$\mathrm{tan}\left(\frac{\pi z}{2}\right)=\mathrm{tan}\left(\frac{\pi \left(z-1\right)}{2}+\frac{\pi }{2}\right)$
$=-\mathrm{cot}\left(\frac{\pi }{2}\left(z-1\right)\right)$
and that therefore
$\underset{z\to 1}{lim}\left(1-z\right)\mathrm{tan}\left(\frac{\pi z}{2}\right)=\underset{z\to 1}{lim}\left(\frac{z-1}{\mathrm{sin}\left(\frac{\pi }{2}\left(z-1\right)\right)}×\mathrm{cos}\left(\frac{\pi }{2}\left(z-1\right)\right)\right)$
$=\frac{\underset{z\to 1}{lim}\mathrm{cos}\left(\frac{\pi }{2}\left(z-1\right)\right)}{\underset{z\to 1}{lim}\frac{\mathrm{sin}\left(\frac{\pi }{2}\left(z-1\right)\right)}{z-1}}$ (since both limits exist)
$=\frac{\mathrm{cos}0}{\frac{\pi }{2}\mathrm{cos}0}$
$=\frac{2}{\pi }$

maul124uk

Let $t\phantom{\rule{0.222em}{0ex}}=\frac{\pi }{2}\left(1-z\right)$
$\underset{z\to 1}{lim}\left(1-z\right)\mathrm{tan}\frac{\pi z}{2}=$
$=\frac{2}{\pi }\underset{t\to 0}{lim}t\mathrm{cot}t=$
$=\frac{2}{\pi }\underset{t\to 0}{lim}\frac{t}{\mathrm{sin}t}\cdot \underset{t\to 0}{lim}\mathrm{cos}t$

Vasquez

$\underset{z\to 1}{lim}\frac{\mathrm{cos}\left(\frac{\pi z}{2}\right)}{1-z}=-\underset{z\to 1}{lim}\frac{\mathrm{cos}\left(\frac{\pi z}{2}\right)-\mathrm{cos}\left(\frac{\pi }{2}\right)}{z-1}=-\mathrm{cos}\left(\frac{\pi x}{2}{\right)}^{\prime }{|}_{x=1}=\frac{\pi }{2}\mathrm{sin}\left(\frac{\pi }{2}\right)=\frac{\pi }{2}$
and
$\underset{z\to 1}{lim}\mathrm{sin}\frac{\pi z}{2}=1$
Combining the above two
$\underset{z\to 1}{lim}\left(1-z\right)\mathrm{tan}\left(\frac{\pi z}{2}\right)=\underset{z\to 1}{lim}\frac{1-z}{\mathrm{cos}\left(\frac{\pi z}{2}\right)}\cdot \underset{z\to 1}{lim}\mathrm{sin}\left(\frac{\pi z}{2}\right)=\frac{2}{\pi }\cdot 1$

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