Limit of $f_{n+1} = \sqrt{12 + f_n}$ with proof by contradiction Consider the following recursive sequence:
$$
\begin{cases}
f_{0}=\sqrt{12}\\
f_{n+1}=\sqrt{12 + f_{n}}
\end{cases}
$$
for $n \geq 0$. How can I prove that this sequence is bounded above by $4$ and
thus its limit is also $4$?
 A: First, being bounded above by $4$ neither guarantees convergence nor, if we knew it converges, that it converges to $4$.
It's easier to show that this converges and then to show that it converges to $4$. To show that it converges, you want to


*

*Show that $f_n$ is increasing.
You can do this directly through induction. It's pretty straightforward.

*Show that $f_n$ is bounded above.
You will have a much easier time by choosing a clever, lenient upper bound. I recommend showing that $f_n < 1 + 2\sqrt{12}$. This number might seem magical. But notice that in the inductive step, we would argue like this:
$$ f_n = \sqrt{12 + f_{n-1}} \leq \sqrt{12 + 2\sqrt{12} + 1} = 1 + \sqrt 12 \lt 1 + 2\sqrt {12}.$$
This choice simplified that inductive argument quite a bit over what it could have been.
Now that you know it converges, you can use the standard argument. Call $L$ the limit. Then we must have $L = \sqrt{12 + L}$. Simplifying, factoring, and throwing out the negative limit shows the actual limit is $4$. $\diamondsuit$
A: Hint: 
By induction  you can prove that the sequence is monoton (strictly) increasing $${ f }_{ 1 }^{ 2 }=12+{ f }_{ 0 }=12+\sqrt { 12 } =12+2\sqrt { 3 } \\ { f }_{ 2 }^{ 2 }=12+12+2\sqrt { 3 } =24+2\sqrt { 3 } \\ { f }_{ 1 }<{ f }_{ 2 }\\ { f }_{ n }<{ f }_{ n+1 }$$
Let say $\lim _{ n\rightarrow \infty  }{ { f }_{ n } } =f$
then $${ f }^{ 2 }-f-12=0$$ $$\left( f-4 \right) \left( f+3 \right) =0$$
so monoton increasing and bounded  above sequence has limit,and limit is $4$
A: Hint:
To prove a recursive sequence is convergent, one strategy is to prove it is non-decreasing and bounded from above, or its non-increasing and bounded from below.
If the sequence is defined with a function $g$,: $f_{n+1}=g(f_n)$, it is  non-decreasing if the graph of $g$ is above  the line $y=x$ on an interval $I$ and all $f_n$ belong to $I$. If the interval $I$ is bounded from above, the sequence will also be bounded from above, hence it will converge.
Furthermore, if $g$ is continuous, the limit $\ell$ will satisfy the equation $\; \ell=g(\ell)$, in other words, the limit will be a fixed point of $g$. If there is only one fixed point, it will be the limit. If there are several fixed  points, you will have to decide (and prove) which fixed point is the limit.
