Hi,I'm trying to prove that [tex]\lim_{x\to 0}\Gamma(x) =

[Froskoy]

Hi,

I'm trying to prove that \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.

 

[aesir]

Hi,

I'm trying to prove that \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?

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.

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.

You are on the right track with the recurrence relation: can you find an expression for \Gamma(1+x) when x\to0?

 

[micromass]

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.

 

[JJacquelin]

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

 

[Froskoy]

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