How can this give an exact result? I was studying about solving expressions of the form $\sqrt{6+\sqrt{6+\sqrt {6+\cdots}}}$
I know how to solve this, but how can this equation have a constant variable if it is continued indefinitely?
I know upto limits and basic derivatives. Thanks
 A: Define a sequence $a_1 = \sqrt 6, a_2 = \sqrt{6 + \sqrt{6}} = \sqrt{6 + a_1}$ and recursively $a_{n+1} = \sqrt{6 + a_n}$ for all $n \geq 1$.
Now show that the sequence $(a_n)$ is bounded above and increasing. Then the sequence has a limit. 
Call that limit $L$. As $L = \lim_{n\to\infty} a_{n+1} = \lim_{n\to\infty} \sqrt{6 + a_n}$ we have 
$$L = \sqrt{6 + L}$$
From this you can find the value of $L$.

We can also generalize this to make sense of expressions such as 
$$\sqrt{a + \sqrt{a + \sqrt{a + \cdots}}}$$
A: If we define the sequence $x_n$ through the recurrence relationship $x_{n+1}=\sqrt{6+x_n}$.  There are three cases for values of $x_1$.
CASE $1$: $x_1=3$
Obviously, if $x_1=3$, then $x_n=3$ for all $n>1$.
CASE $2$: $-6\le x_1<3$
If $-6\le x_1<3$, then it is easy to see inductively that $0<x_{n}<3$ for all $n > 1$.  Additionally, it is easy to see that $x_{n+1}-x_n=\sqrt{6+x_n}-x_n=\frac{(3-x_n)(2+x_n)}{\sqrt{6+x_n}+x_n}>0$ and thus $x_n$ increases monotonically.  Therefore we have 
$$\lim_{n\to \infty}x_{n+1}=\lim_{n\to \infty}x_n=\sqrt{6+\lim_{n\to \infty}x_n}\implies \lim_{n\to \infty}x_n=3$$
CASE $3$: $3<x_1$
If $3<x_1$, then it is easy to see inductively that $0<x_{n}>3$ for all $n\ge 1$.  Additionally, it is easy to see that $x_{n+1}-x_n=\sqrt{6+x_n}-x_n=\frac{(3-x_n)(2+x_n)}{\sqrt{6+x_n}+x_n}<0$ and thus $x_n$ decreases monotonically.  
Therefore we have $$\lim_{n\to \infty}x_{n+1}=\lim_{n\to \infty}x_n=\sqrt{6+\lim_{n\to \infty}x_n}\implies \lim_{n\to \infty}x_n=3$$
