How do I evaluate limz→1(1−z)tan⁡πz2 I tried using the identity, tan⁡x2=1−cos⁡xsin⁡x to simplify this to: limz→1(1−z)sin⁡πz1+cos⁡πz

Note that tan⁡(πz2)=tan⁡(π(z−1)2+π2) =−cot⁡(π2(z−1)) and that therefore limz→1(1−z)tan⁡(πz2)=limz→1(z−1sin⁡(π2(z−1))×cos⁡(π2(z−1))) =limz→1cos⁡(π2(z−1))limz→1sin⁡(π2(z−1))z−1 (since both limits exist) =cos⁡0π2cos⁡0 =2π

Let t=π2(1−z) limz→1(1−z)tan⁡πz2= =2πlimt→0tcot⁡t= =2πlimt→0tsin⁡t⋅limt→0cos⁡t

limz→1cos⁡(πz2)1−z=−limz→1cos⁡(πz2)−cos⁡(π2)z−1=−cos⁡(πx2)′|x=1=π2sin⁡(π2)=π2andlimz→1sin⁡πz2=1Combining the above twolimz→1(1−z)tan⁡(πz2)=limz→11−zcos⁡(πz2)⋅limz→1sin⁡(πz2)=2π⋅1

Note that tan⁡(πz2)=tan⁡(π(z−1)2+π2) =−cot⁡(π2(z−1)) and that therefore limz→1(1−z)tan⁡(πz2)=limz→1(z−1sin⁡(π2(z−1))×cos⁡(π2(z−1))) =limz→1cos⁡(π2(z−1))limz→1sin⁡(π2(z−1))z−1 (since both limits exist) =cos⁡0π2cos⁡0 =2π

Let t=π2(1−z) limz→1(1−z)tan⁡πz2= =2πlimt→0tcot⁡t= =2πlimt→0tsin⁡t⋅limt→0cos⁡t

limz→1cos⁡(πz2)1−z=−limz→1cos⁡(πz2)−cos⁡(π2)z−1=−cos⁡(πx2)′|x=1=π2sin⁡(π2)=π2andlimz→1sin⁡πz2=1Combining the above twolimz→1(1−z)tan⁡(πz2)=limz→11−zcos⁡(πz2)⋅limz→1sin⁡(πz2)=2π⋅1

Solved: "How do I evaluate limz→1(1−z)tan⁡πz2 I tried using..."

High School
Geometry
Trigonometry
Trigonometric equations and identities

Brock Brown

Answered question

2022-01-02

How do I evaluate

lim_{z \to 1} (1 - z) \tan \frac{π z}{2}

I tried using the identity,

\tan \frac{x}{2} = \frac{1 - \cos x}{\sin x}

to simplify this to:

lim_{z \to 1} (1 - z) \frac{\sin π z}{1 + \cos π z}

Answer & Explanation

psor32

Beginner2022-01-03Added 33 answers

Note that

\tan (\frac{π z}{2}) = \tan (\frac{π (z - 1)}{2} + \frac{π}{2})

= - \cot (\frac{π}{2} (z - 1))

and that therefore

lim_{z \to 1} (1 - z) \tan (\frac{π z}{2}) = lim_{z \to 1} (\frac{z - 1}{\sin (\frac{π}{2} (z - 1))} \times \cos (\frac{π}{2} (z - 1)))

= \frac{lim_{z \to 1} \cos (\frac{π}{2} (z - 1))}{lim_{z \to 1} \frac{\sin (\frac{π}{2} (z - 1))}{z - 1}}

(since both limits exist)

= \frac{\cos 0}{\frac{π}{2} \cos 0}

= \frac{2}{π}

maul124uk

Beginner2022-01-04Added 35 answers

Let

t = \frac{π}{2} (1 - z)

lim_{z \to 1} (1 - z) \tan \frac{π z}{2} =

= \frac{2}{π} lim_{t \to 0} t \cot t =

= \frac{2}{π} lim_{t \to 0} \frac{t}{\sin t} \cdot lim_{t \to 0} \cos t

Vasquez

Expert2022-01-08Added 669 answers

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

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