How to calculate limit $\lim_{x\to\infty}\left(\lfloor x\rfloor+\frac{1}{2}\right)\ln x-\ln(\lfloor x\rfloor!)$

Question

While uses plot on www.wolframalpha.com, i found a problem: The function $$f(x)=\left(\lfloor x\rfloor+\frac{1}{2}\right)\ln x-\ln(\lfloor x\rfloor!)$$ approximate the function $$g(x)=x-\frac{1}{2}\ln\pi$$

Can you prove it?

Answer 1

Use Stirling's formula:

$$x!\sim \sqrt{2\pi x}\left( \frac{x}{e}\right)^x$$

$$\begin{align} \lim_{x\to\infty} f(x)&=(x+\frac{1}2)\ln(x)-\frac{1}{2}\ln(2\pi x)-x\ln(x)+x\\ \\ \lim_{x\to\infty} f(x)&=x-\frac{1}{2}\ln(2\pi) \end{align}$$

Note in your plot (and in your post), it should be $g(x)=x-\frac{1}{2}\ln(2\pi)$

Here is the updated plot