How do I evaluate \lim_{z \to 1}(1-z)\tan \frac{\pi z}{2} I tried

Brock Brown 2022-01-02 Answered
How do I evaluate limz1(1z)tanπz2
I tried using the identity, tanx2=1cosxsinx to simplify this to:
limz1(1z)sinπz1+cosπz
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Expert Answer

psor32
Answered 2022-01-03 Author has 33 answers
Note that
tan(πz2)=tan(π(z1)2+π2)
=cot(π2(z1))
and that therefore
limz1(1z)tan(πz2)=limz1(z1sin(π2(z1))×cos(π2(z1)))
=limz1cos(π2(z1))limz1sin(π2(z1))z1 (since both limits exist)
=cos0π2cos0
=2π
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maul124uk
Answered 2022-01-04 Author has 35 answers
Let t=π2(1z)
limz1(1z)tanπz2=
=2πlimt0tcott=
=2πlimt0tsintlimt0cost
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Vasquez
Answered 2022-01-08 Author has 460 answers

limz1cos(πz2)1z=limz1cos(πz2)cos(π2)z1=cos(πx2)|x=1=π2sin(π2)=π2
and
limz1sinπz2=1
Combining the above two
limz1(1z)tan(πz2)=limz11zcos(πz2)limz1sin(πz2)=2π1

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