Question

Find \(lt_{p\rightarrow\infty}\frac{p^{\frac{1}{2}}.p!}{\frac{1}{2}.\frac{3}{2}…(p+\frac{1}{2})}\)

(a) √π

(b) ∞

(c) √^π⁄2

(d) 0

I had been asked this question in final exam.

This key question is from Indeterminate Forms in portion Differential Calculus of Engineering Mathematics

Answer 1

The correct answer is (c) √^π⁄2

Easiest explanation: Using the Gauss definition of the Gamma function we have

\(\tau(x)=lt_{p\rightarrow\infty}\frac{p^x.p!}{x.(x+1)…(x+p)}\)

Where τ(x) is the Gamma function Using formula

\(\tau(x) \times \tau(1-x)=\frac{\pi.x}{sin(\pi.x)}\)

put x=1/2 to get

\((\tau(\frac{1}{2}))^2=\frac{\pi}{2.sin(\frac{\pi}{2})}=\frac{\pi}{2}\)

\(\tau(\frac{1}{2})=\sqrt{\frac{\pi}{2}}\)