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

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