Limit of $\frac{(x^x)}{(x!)}$ as $x$ approaches infinity So, I'm learning limits right now in calculus class.
When $x$ approaches infinity, what does this expression approach?
$$\frac{(x^x)}{(x!)}$$
Why?  Since, the bottom is $x!$, doesn't it mean that the bottom goes to zero faster, therefore the whole thing approaches 0?
 A: Hint
$$\frac{x^x}{x!} > \frac{x}{1}$$
A: The limit is $\infty$, think about it this way:
On the top, $$x^x=\underbrace{x\times x\times\cdots\times x}_{x\text{ number of }x's}$$
On the bottom, $$x!=x\times(x-1)\times\cdots\times 1$$
Which is is bigger?  Can you tell?
A: I figured I'd go for a more "proper" proof. Notice that, if we let $a_n = n^n/n!$ we can write that
$$
a_{n+1}-a_n = \frac{(n+1)^{n+1}}{(n+1)!}-\frac{n^n}{n!} = \frac{(n+1)^{n+1}-(n+1)n^n}{(n+1)!}
$$
which can then be written as
$$
\frac{(n+1)^n-n^n}{n!}
$$
Now, using the binomial theorem, the first term remaining in the numerator is
$$
\binom{n}{1}n^n\cdot1^n = n^{n+1}
$$
And all terms in the numerator are positive. Therefore, we have
$$
a_{n+1}-a_n > \frac{n^{n+1}}{n!} = na_n
$$
and so
$$
a_{n+1} > (n+1)a_n
$$
Therefore, as $a_1=1$, we can clearly see that (for $n>1$)
$$
a_n > \frac{n!}2
$$
and thus
$$
\lim_{n\to \infty} a_n \to \infty
$$
A: Here's $10^{10}$:
$$10 \cdot 10 \cdot 10\cdot 10\cdot 10\cdot 10\cdot 10\cdot 10\cdot 10\cdot 10$$
Here's $10!$:
$$10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$$
Which one is bigger?  Carry that thought out for larger and larger numbers, and you'll see the answer.
A: This ratio grows at the rate $e^x \sqrt{2 \pi x}(1+o(1))$ which tends to infininty as $x \to \infty$
A: You may be thinking of the fact that $e^x/x!\to 0$. The difference is that there the base of the exponent is fixed, while in this problem, the base of the exponent grows.
A: There is a theorem, which states that, if the reciprocal approaches zero, then the original expression approches $+\infty$. That might me a strategy in this case.
A: You may use Stirlings approximation to do this.
Let $$y = x^x/x! $$, then $$ln(y) = ln((x^x/x!)) = ln(x^x) - ln(x!)$$
The last term on the right hand side can be expanded as $$ ln(x!) = xln(x) - x $$ for large n. As a side note, this is a typical expansion used in physics.
Substituting back and taking limits, $$ \lim_{x \to +\infty} ln(y) = xln(x) - (xln(x) -x )$$
$$ \lim_{x \to +\infty}\Rightarrow ln(y) = \lim_{x \to +\infty}x = \infty $$.
Stirlings approximation induces an approximation of O(log(x)) which goes as $$ \sqrt{2\pi x}$$ in the first term. If this is included and reworked this is will be result as @alex suggested
