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 \(\displaystyle\lim_{{{z}\to{1}}}{\left({1}-{z}\right)}{\tan{{\frac{{\pi{z}}}{{{2}}}}}}\)
I tried using the identity, \(\displaystyle{\tan{{\frac{{{x}}}{{{2}}}}}}={\frac{{{1}-{\cos{{x}}}}}{{{\sin{{x}}}}}}\) to simplify this to:
\(\displaystyle\lim_{{{z}\to{1}}}{\left({1}-{z}\right)}{\frac{{{\sin{\pi}}{z}}}{{{1}+{\cos{\pi}}{z}}}}\)

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Expert Answer

psor32
Answered 2022-01-03 Author has 2192 answers
Note that
\(\displaystyle{\tan{{\left({\frac{{\pi{z}}}{{{2}}}}\right)}}}={\tan{{\left({\frac{{\pi{\left({z}-{1}\right)}}}{{{2}}}}+{\frac{{\pi}}{{{2}}}}\right)}}}\)
\(\displaystyle=-{\cot{{\left({\frac{{\pi}}{{{2}}}}{\left({z}-{1}\right)}\right)}}}\)
and that therefore
\(\displaystyle\lim_{{{z}\to{1}}}{\left({1}-{z}\right)}{\tan{{\left({\frac{{\pi{z}}}{{{2}}}}\right)}}}=\lim_{{{z}\to{1}}}{\left({\frac{{{z}-{1}}}{{{\sin{{\left({\frac{{\pi}}{{{2}}}}{\left({z}-{1}\right)}\right)}}}}}}\times{\cos{{\left({\frac{{\pi}}{{{2}}}}{\left({z}-{1}\right)}\right)}}}\right)}\)
\(\displaystyle={\frac{{\lim_{{{z}\to{1}}}{\cos{{\left({\frac{{\pi}}{{{2}}}}{\left({z}-{1}\right)}\right)}}}}}{{\lim_{{{z}\to{1}}}{\frac{{{\sin{{\left({\frac{{\pi}}{{{2}}}}{\left({z}-{1}\right)}\right)}}}}}{{{z}-{1}}}}}}}\) (since both limits exist)
\(\displaystyle={\frac{{{\cos{{0}}}}}{{{\frac{{\pi}}{{{2}}}}{\cos{{0}}}}}}\)
\(\displaystyle={\frac{{{2}}}{{\pi}}}\)
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maul124uk
Answered 2022-01-04 Author has 833 answers
Let \(\displaystyle{t}\:={\frac{{\pi}}{{{2}}}}{\left({1}-{z}\right)}\)
\(\displaystyle\lim_{{{z}\to{1}}}{\left({1}-{z}\right)}{\tan{{\frac{{\pi{z}}}{{{2}}}}}}=\)
\(\displaystyle={\frac{{{2}}}{{\pi}}}\lim_{{{t}\to{0}}}{t}{\cot{{t}}}=\)
\(\displaystyle={\frac{{{2}}}{{\pi}}}\lim_{{{t}\to{0}}}{\frac{{{t}}}{{{\sin{{t}}}}}}\cdot\lim_{{{t}\to{0}}}{\cos{{t}}}\)
0
Vasquez
Answered 2022-01-08 Author has 9499 answers

\(\lim_{z \to 1} \frac{\cos(\frac{\pi z}{2})}{1-z}=-\lim_{z \to 1} \frac{\cos(\frac{\pi z}{2})-\cos(\frac{\pi}{2})}{z-1}=-\cos(\frac{\pi x}{2})' \Bigg|_{x=1}=\frac{\pi}{2}\sin (\frac{\pi}{2})=\frac{\pi}{2}\)
and
\(\lim_{z \to 1} \sin \frac{\pi z}{2}=1\)
Combining the above two
\(\lim_{z \to 1}(1-z)\tan(\frac{\pi z}{2})=\lim_{z \to 1} \frac{1-z}{\cos(\frac{\pi z}{2})} \cdot \lim_{z \to 1} \sin(\frac{\pi z}{2})=\frac{2}{\pi} \cdot 1\)

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