Why does $\lim\limits_{x\to\infty}(x!)^{1/x}\neq 1?$ Why does $\lim\limits_{x\to\infty}(x!)^{1/x}\neq 1?$
As far as I know, anything to the power of $0$ is $1$.
We have a factorial raised to $1/\infty=0$, but the limit is not $1$? I don't even know what the limit is. But it seems like infinity? Why is this?

 A: That doesn't follow: $1/x$ is not the same as $0$ even for very large values of $x$. Keep in mind that "the limit as $x$ goes to infinity" is not the same thing as "plug in infinity wherever you see $x$". It is about how the function is behaving for larger and larger $x$. No matter how large $x$ gets, $1/x$ is still positive and as such we can find a (possibly huge) real number $\alpha$ such that $\alpha^{1/x}$ is as big as we want. For example, you will agree that
$$\lim_{x \to \infty} (a^x)^{1/x} = a.$$
Combining this with the fact that $x! > a^x$ for any $a$ and sufficiently large $x$, it shouldn't be too surprising that
$$\lim_{x \to \infty} (x!)^{1/x} = \infty.$$
A: Method (1). Prove the following: If $f(x)\to \infty$ as $x\to \infty$ then $G(n)=\frac {1}{n}\sum_{j=1}^nf(j)\to \infty$ as $n\to \infty.$
With $f(x)=\ln x$ we have $G(n)=\ln (n!^{(1/n)}).$
Method (2). $\ln x$ is monotonic increasing. So for $n\geq 2$ we have $\ln n>\int_{n-1}^n \ln x\;dx  .$ So $$\ln n!=\sum_{j=2}^n \ln j>\sum_{j=2}^n\int_{j-1}^j\ln x\;dx=\int_1^n \ln x \;dx=n(\ln n)-n+1.$$ Therefore for n\geq 2 we have  $$\ln (n!^{(1/n)})=\frac {1}{n}\ln n!>\frac {1}{n}(n (\ln n)-n+1)=(\ln n)-1+\frac {1}{n}.$$
A: A useful inequality is
$n! > (n/e)^n$.
This can be proved by induction
from
$(1+1/n)^n < e$.
From this
$(n!)^{1/n} > n/e$
and this last is unbounded.
A: As Salahamam_ Fatima commented, think about Stirling approximation.
$$y=(x!)^{1/x}\implies \log(y)=\frac 1x \log(x!)$$ Now, using Stirling approximation $$\log(x!)=x (\log (x)-1)+\frac{1}{2} \left(\log (2 \pi )+\log
   \left({x}\right)\right)+O\left(\frac{1}{x}\right)$$ $$\log(y)=(\log (x)-1)+\frac{\log (2 \pi )+\log \left({x}\right)}{2 x}+O\left(\frac{1}{x^2}\right)$$ and now,  $$y=e^{\log(y)}=\frac{x}{e}+\frac{\log (2 \pi )+\log \left({x}\right)}{2 e}+O\left(\frac{1}{x}\right)$$
