$\lim_{n\to\infty} \frac{10^n}{\sqrt{(n+1)!} + \sqrt{n!}}\,.$

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}

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 $$0 \lt \frac{a^n}{n!} = \frac{a}{1}\frac{a}{2} \cdots \frac{a}{k} \frac{a}{k+1}\cdots \frac{a}{n} \leqslant \frac{a}{1}\frac{a}{2} \cdots \frac{a}{k} \left(\frac{a}{k+1}\right)^{n-k} \to 0$$

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 $\lim_{n\to\infty} a_n = 0$.