# Hi,I'm trying to prove that $$\lim_{x\to 0}\Gamma(x) = 1. Jan 4, 2012 ### Froskoy Hi, I'm trying to prove that [tex]\lim_{x\to 0}\Gamma(x) = \infty$$ but I'm having a little trouble understanding.

Using the recurrence relation $$\Gamma(x)=\frac{\Gamma(x+1)}{x}$$ you can see that as the denominator tends towards zero, you'd expect the whole thing to get bigger and bigger...
but doesn't that mean that the numerator will get bigger and bigger as well because the next value of $\Gamma(x)$ will be bigger as well so it won't tend to infinity?

So how do you show this formally? Is there some mathematical was of quantifying how quickly the function tends towards infinity?

With many thanks,

Froskoy.

2. Jan 5, 2012

### aesir

Re: Gamma(0)

No, the next value is $\Gamma(1)=\lim_{x\to 0}x \Gamma(x)$: using the previous limit would give $0\times\infty$ which is undetermined.
You are on the right track with the recurrence relation: can you find an expression for $\Gamma(1+x)$ when $x\to0$?

3. Jan 5, 2012

### micromass

Re: Gamma(0)

I think it's easier using the definition of the Gamma function:

$$\lim_{z\rightarrow 0}{ \int_0^{+\infty}{ e^{-t}t^zdt } }$$

Now use the theorem of dominated convergence to interchange limit and integral.

4. Jan 5, 2012

### JJacquelin

Re: Gamma(0)

x -> 0
Gamma(1+x) -> Gamma(1) = 1
x = Gamma(1+x)/x -> 1/x -> infinity

5. Jan 6, 2012

### Froskoy

Re: Gamma(0)

Thanks everyone! It is much clearer now and I have done it both ways.