Help solving a not so difficult limit? Giving that $(a_{n})$ is an arithmetic series of ratio $r$ , why is
$$ \lim_{n\to\infty}\frac{1}{\frac{\sqrt{a_{n+1}}}{\sqrt{r}}\arctan \frac{\sqrt{r}}{\sqrt{a_{n+1}}}}\cdot \frac{1}{1 + \frac{\sqrt{a_{n+2}}}{\sqrt{a_n+1}}} = \frac{1}{2} ?$$
This is part of an exercise I'm trying to understand, and this is what I yet have to find an explanation for.
 A: The first factor converges to $1$ because $$\lim\limits_{x\to\infty}\frac{x}{\arctan x}=1,$$ while as for the second factor we have $$\require\cancel \lim_{n\to\infty}\frac{1}{1 + \frac{\sqrt{a_{n+2}}}{\sqrt{a_n+1}}}=\lim_{n\to\infty} \frac{\sqrt{a_n+1}}{\sqrt{a_{n}+2r}+\sqrt{a_n+1}}=\lim_{n\to\infty} \frac{\cancel{\sqrt{a_n}}\sqrt{1+1/a_n}}{\cancel{\sqrt{a_n}}(\sqrt{1+2r/a_n}+\sqrt{1+1/a_n})}=\frac{1}{2}.$$
A: You have to solve this limit by taking each ratio in it separately. In the first you'll have to make a substitution: Note $\frac{\sqrt{a_{n+1}}}{\sqrt{r}}=x$, hence n $\rightarrow \infty \Rightarrow x\rightarrow \infty$ . Thus, you will get $\lim_{x\rightarrow \infty}\frac{1}{xarctan\frac{1}{x}}$. In this case at the denominator you will have $\infty*0$ ,so we will write that limit somehow else: ,now applying L'Hospital rules we get (after some computations and simplifications) $\lim_{x\rightarrow \infty}\frac{1}{1+\frac{1}{x^2}}$ which is 1. So, the limit of the first ratio is 1. One can easily show that the limit of the second ratio is $\frac{1}{2}$ ,since we write the ratio $\frac{\sqrt{a_{n+2}}}{\sqrt{a_{n+1}}}$ as $\frac{\sqrt{(n+1)r+a_1}}{\sqrt{nr+a_1}}$ and this limit is 1 so the second limit is $\frac{1}{2}$. Henceforth, the requested limit is $\frac{1}{2}$.
