$\lim_{n\to\infty} \frac{10^n}{\sqrt{(n+1)!} + \sqrt{n!}}\,.$ $\lim_{n\to\infty} \frac{10^n}{\sqrt{(n+1)!} + \sqrt{n!}}\,.$
I used the ratio criterion for the calculation and I got to this, can I say now that it is zero or is it still an undefined expression?
$\frac{10+(10/\sqrt{(n+1)})}{\sqrt{(n+2)}+1}$
10/inf < 1 ----> lim an=0

I would like to completely eliminate the doubt regarding why " n! is 'stronger', then any ns power", with the aid of Sterling's approximation, and solve the question even without using the ratio test.....
$\frac{10^n}{\sqrt{(n+1)!} + \sqrt{n!}} \lt \frac{10^n}{2\sqrt{n!}}=\frac{1}{2}\sqrt{\frac{100^n}{n!}}$
This inequality true because the denominator of right term is smaller than left term...
By the Sterling's approximation, we have
$n!\sim \frac{\sqrt{2\pi n}n^n}{e^n}$
Hence, at the limit to infinity, we can write
$\frac{1}{2}\sqrt{\frac{100^n}{n!}}=\frac{1}{2}\sqrt{\frac{100^n}{\sqrt{2\pi n}~n^n}}=\frac{1}{2}\sqrt{\frac{1}{\sqrt{2\pi n}}~.({\frac{100}{n})}^n}=0$,  as the limit$\frac{100}{n}$ goes to 0..
Here the important fact is then 100, can be replaced by any real finite number..., hence proving why n! is stronger
