Find if the recursive sequence is limited and monototonic. 
I'm back with another question about sequences which we at the moment we are tought at school.
Let $a_{1} = 5$ and recursive formula $a_{n+1} = 5\sqrt{a_{n}-1} -3$ .
I think I did it right for finding if the sequence is monotoniclly growing.
$a_{n+1} \ge a_{n}$ 
After a few steps I came with a quadratic function and by calculating when this function $\le 0$, I got result that $a_{n} \in[2,17]$

I don't know if it's mathematically correct now to just say that, yes the upper limit is 17 and lower limit is 5 and yes, it is monotonically growing.
Anyone can help me with this?
Thanks

We prove by induction that the sequence is increasing.
We can check directly that $a_2\gt a_1$. Now suppose that we know that for a certain $k$ we have $a_{k+1}\gt a_k$. Then 
$$a_{k+2}=5\sqrt{a_{k+1}-1}-3\gt 5\sqrt{a_k-1}-3=a_{k+1}.$$
To show the sequence has a limit, it is enough now to show that the sequence is
bounded  above. We show by induction that $a_n\lt 101$ for all $n$. This is certainly true for $n=1$. Suppose it is true for $n=k$. We show it is true  for $n=k+1$.  We have
$$a_{k+1}=5\sqrt{a_k-1}-3\lt 5\sqrt{100}-3=47\lt 101.$$
Remark: Now that we know the sequence has a limit $L$, it is not hard to calculate $L$. For 
$$\lim_{n\to\infty}a_{n+1}=5\lim_{n\to\infty}\sqrt{a_n-1}-3.$$
Thus $L=5\sqrt{L-1}-3$. A little manipulation gives us a quadratic equation in $L$.
