Evaluating the limit of a gamma function

Question

I have the following to start:

$$F(x)=x^{b-a}\frac{\Gamma(x+a+1)}{\Gamma(x+b+1)}$$

And I'm trying to evaluate:

$$\lim_{x\rightarrow\infty}F(x)$$

I have simplified this to yield the same outcome as

$$e^{b-a} \lim_{x\rightarrow\infty}x^{b-a}\frac{(x+a)^{x+a}}{(x+b)^{x+b}}$$

But I am totally stuck from here.

Any input on how to evaluate the initial limit or how to evaluate the limit I have simplified to would be excellent. I know that using Stirling's approximation is useful in finding the limit of the initial problem and that is in fact how I reached my simplification.

Thanks in advance.

Answer 1

We have $$e^{b-a}\lim_{x\rightarrow\infty}x^{b-a}\frac{\left(x+a\right)^{x+a}}{\left(x+b\right)^{x+b}}=e^{b-a}\lim_{x\rightarrow\infty}\frac{\left(x+a\right)^{a}}{x^{a}}\frac{x^{b}}{\left(x+b\right)^{b}}\frac{\left(x+a\right)^{x}}{\left(x+b\right)^{x}}=e^{b-a}\lim_{x\rightarrow\infty}\frac{\left(1+\frac{a}{x}\right)}{\left(1+\frac{b}{x}\right)^{b}}^{a}\left(\frac{x+a}{x+b}\right)^{x}=e^{b-a}\lim_{x\rightarrow\infty}\left(1+\frac{a-b}{x+b}\right)^{x}=e^{b-a}e^{a-b}=1.$$

Answer 2

You can treat this as a rational fraction by simplifying out all the $x$, whose powers do cancel out: you get $$F(x) \sim (1+a/x)^{x+a} (1+b/x)^{-x-b}.$$ We would now simplify this by using the limited series expansion of $(1+t)^u$, but here $u$ is a function of $x$, so we must first show that this is valid. Namely, we have $$\begin{split} (1+a/x)^{x+a} &= \exp ((x+a) \log (1+a/x)) = \exp ((x+a) (a/x+O(x^{-2})) \\ &= \exp (a + O(1/x))\end{split},$$ so that unless I am mistaken, $$F(X) \sim e^{b-a} \exp(a+O(1/x)) \exp(-b+O(1/x)) \rightarrow 1$$ as $x \rightarrow +\infty$.

Answer 3

Just a disclaimer that I've learned everything I know about asymptotic expansions from reading random solutions on this very site.

Since (from the wikipedia entry on the Gamma function)

$$\lim_{n\rightarrow \infty} \frac{\Gamma(n+\alpha)}{\Gamma(n)n^{\alpha}} = 1, \quad \alpha \in \mathbb{R}.$$

Write

\begin{align*} \lim_{x\rightarrow} F(x)&= \lim_{x\rightarrow \infty} x^{b-a}\frac{\Gamma(x+a+1)}{\Gamma(x+b+1)}\\ &=\lim_{n\rightarrow \infty} n^{b-a}\frac{\Gamma(n+a+1)}{\Gamma(n+b+1)}\\ &= \lim_{n\rightarrow\infty} n^{b-a}\frac{\Gamma(n)n^{a+1}}{\Gamma(n)n^{b+1}}\\ &= n^{b-a}n^{a+1}n^{-b-1}\\ &= 1. \end{align*}

Answer 4

Continuing in the same spirit as Circonflexe, we could show how the limit is approached.

To the next order, we have $$\Big(1+\frac ax\Big)^{x+a}=e^a+\frac{a^2 e^a}{2 x}+O\left(\left(\frac{1}{x}\right)^2\right)$$ from which $$F(x)=x^{b-a}\frac{\Gamma(x+a+1)}{\Gamma(x+b+1)}=1+\frac{(a-b) (a+b+1)}{2 x}+O\left(\left(\frac{1}{x}\right)^2\right)$$

With regard to the approximation in the Wikipedia page, we could show, using the same way that, for large values of $n$, $$\frac{\Gamma(n+a)}{\Gamma(n)~n^a}=1+\frac{(a-1) a}{2 n}+\frac{(a-2) (a-1) a (3 a-1)}{24 n^2}+O\left(\left(\frac{1}{n}\right)^3\right)$$