$\lim_{n\to\infty} \frac{10^n}{\sqrt{(n+1)!} + \sqrt{n!}}\,.$ $\lim_{n\to\infty} \frac{10^n}{\sqrt{(n+1)!} + \sqrt{n!}}\,.$
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
 A: Using sandwich theorem is enough for this problem:
\begin{align}
0\le \lim_{n\to\infty} \frac{10^n}{\sqrt{(n+1)!} + \sqrt{n!}}&\le\lim_{n\to\infty}\dfrac{10^n}{\sqrt{n!}+\sqrt{n!}}\\
&=\dfrac12\times\lim_{n\to\infty}\sqrt{\dfrac{100^n}{n!}}\\
&=\dfrac12\times\sqrt{\lim_{n\to\infty}\left(\dfrac{100^n}{n!}\right)}\\
&=0
\end{align}
A: Using
$$\frac{10^n}{\sqrt{(n+1)!} + \sqrt{n!}} \lt \frac{10^n}{2\sqrt{n!}}=\frac{1}{2}\sqrt{\frac{100^n}{n!}}$$
we come to limit $\frac{a^n}{n!}$ for $a \gt 1$. Of course exists natural $k=\lfloor a\rfloor \leqslant a \leqslant \lfloor a\rfloor +1$. So we have
$$0 \lt \frac{a^n}{n!} = \frac{a}{1}\frac{a}{2} \cdots \frac{a}{k} \frac{a}{k+1}\cdots \frac{a}{n} \leqslant \frac{a}{1}\frac{a}{2} \cdots \frac{a}{k} \left(\frac{a}{k+1}\right)^{n-k} \to 0$$
A: Hint: No, it's not undefined, since the numerator goes to 11.
Just explaining a bit about the above answer:
$$100^n \leq  n!$$
and because we can see that:
$$n \cdot log(100) < log(n!) = \sum_{j = 0 }^{n} log(j)$$
for some $j = 100$, from this point the sums on the right will be bigger and overcome the difference, this is, on the left you are summing each time $log(100)$, but at some moment a gap will appear as we said, and this can be proved formally.
A: As you have pointed out,
$$\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = \lim_{n\to\infty} \frac{10+\frac{10}{\sqrt{n+1}}}{\sqrt{n+2}+1}$$ So,
$$\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = \lim_{n\to\infty} (\frac{10}{\sqrt{n+2}+1} + \frac{\frac{10}{\sqrt{n+1}}}{\sqrt{n+2}+1}) = \lim_{n\to\infty} \frac{10}{\sqrt{n+2}+1} + \lim_{n\to\infty} \frac{10}{(\sqrt{n+1})(\sqrt{n+2}+1)} = 0$$ since both limits exist. So, since $$\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = 0 < 1$$ you can say that $\lim_{n\to\infty} a_n = 0$.
A: 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
