I tried using the identity,

Brock Brown
2022-01-02
Answered

How do I evaluate $\underset{z\to 1}{lim}(1-z)\mathrm{tan}\frac{\pi z}{2}$

I tried using the identity,$\mathrm{tan}\frac{x}{2}=\frac{1-\mathrm{cos}x}{\mathrm{sin}x}$ to simplify this to:

$\underset{z\to 1}{lim}(1-z)\frac{\mathrm{sin}\pi z}{1+\mathrm{cos}\pi z}$

I tried using the identity,

You can still ask an expert for help

psor32

Answered 2022-01-03
Author has **33** answers

Note that

$\mathrm{tan}\left(\frac{\pi z}{2}\right)=\mathrm{tan}(\frac{\pi (z-1)}{2}+\frac{\pi}{2})$

$=-\mathrm{cot}\left(\frac{\pi}{2}(z-1)\right)$

and that therefore

$\underset{z\to 1}{lim}(1-z)\mathrm{tan}\left(\frac{\pi z}{2}\right)=\underset{z\to 1}{lim}(\frac{z-1}{\mathrm{sin}\left(\frac{\pi}{2}(z-1)\right)}\times \mathrm{cos}\left(\frac{\pi}{2}(z-1)\right))$

$=\frac{\underset{z\to 1}{lim}\mathrm{cos}\left(\frac{\pi}{2}(z-1)\right)}{\underset{z\to 1}{lim}\frac{\mathrm{sin}\left(\frac{\pi}{2}(z-1)\right)}{z-1}}$ (since both limits exist)

$=\frac{\mathrm{cos}0}{\frac{\pi}{2}\mathrm{cos}0}$

$=\frac{2}{\pi}$

and that therefore

maul124uk

Answered 2022-01-04
Author has **35** answers

Let $t{\textstyle \phantom{\rule{0.222em}{0ex}}}=\frac{\pi}{2}(1-z)$

$\underset{z\to 1}{lim}(1-z)\mathrm{tan}\frac{\pi z}{2}=$

$=\frac{2}{\pi}\underset{t\to 0}{lim}t\mathrm{cot}t=$

$=\frac{2}{\pi}\underset{t\to 0}{lim}\frac{t}{\mathrm{sin}t}\cdot \underset{t\to 0}{lim}\mathrm{cos}t$

Vasquez

Answered 2022-01-08
Author has **460** answers

and

Combining the above two

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