Ghetto Stirling Approximation?

Calculus Level 2

Determine the value of $\lfloor{1000a}\rfloor$

where $a = \displaystyle\lim_{x\rightarrow \infty} \frac{x^{x+0.5}}{x!e^{x}}$

1 solution

X X
Sep 15, 2018

Use Stirling's Formula, $n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n$ .

So $\displaystyle\lim_{x\rightarrow \infty} \frac{x^{x+0.5}}{x!e^{x}}$

$=\displaystyle\lim_{x\rightarrow \infty} \frac{x^{x+0.5}}{e^x\sqrt{2 \pi x}\left(\frac{x}{e}\right)^x}$

$=\dfrac1{\sqrt{2\pi}}\approx 0.3989...$