UPSC 2018 Maths Optional Paper 1 Q1c — Step-by-Step Solution

10 marks · Section A

Indeterminate forms · Calculus · asked 4× in 13 yrs · Read the full method →

Question

Determine if $\displaystyle\lim_{z\to1}(1-z)\tan\frac{\pi z}{2}$ exists or not. If the limit exists, then find its value.

$0\cdot\infty$ form; shift $z=1+h$ , use $\tan(\frac\pi2+\theta)=-\cot\theta$ , then $\sin\theta/\theta\to1$ .

As $z\to1$ , $(1-z)\to0$ while $\tan\frac{\pi z}{2}\to\tan\frac{\pi}{2}=\pm\infty$ . The product is of the indeterminate form $0\cdot\infty$ .

Let $z=1+h$ with $h\to0$ . Then $1-z=-h$ and

\tan\frac{\pi z}{2}=\tan\!\left(\frac{\pi}{2}+\frac{\pi h}{2}\right)=-\cot\frac{\pi h}{2},

using $\tan\!\left(\frac\pi2+\theta\right)=-\cot\theta$ . Hence

(1-z)\tan\frac{\pi z}{2}=(-h)\!\left(-\cot\frac{\pi h}{2}\right)=h\cot\frac{\pi h}{2}=h\cdot\frac{\cos\frac{\pi h}{2}}{\sin\frac{\pi h}{2}}.

Write $\sin\frac{\pi h}{2}=\frac{\pi h}{2}\cdot\dfrac{\sin\frac{\pi h}{2}}{\frac{\pi h}{2}}$ . Then

h\cot\frac{\pi h}{2}=\frac{h\cos\frac{\pi h}{2}}{\frac{\pi h}{2}\cdot\frac{\sin(\pi h/2)}{\pi h/2}}=\frac{2}{\pi}\cdot\frac{\cos\frac{\pi h}{2}}{\dfrac{\sin(\pi h/2)}{\pi h/2}}.

As $h\to0$ : $\cos\frac{\pi h}{2}\to1$ and $\dfrac{\sin(\pi h/2)}{\pi h/2}\to1$ . The two‑sided limit exists and equals $\dfrac2\pi$ .

\boxed{\;\lim_{z\to1}(1-z)\tan\frac{\pi z}{2}=\frac{2}{\pi}.\;}