$\sqrt{2 + \sqrt{2+\sqrt{2+\sqrt2...)}}}$.
Pretty classic question, I think - and the limit is equal to 2.
But how do I prove this rigorously? An epsilon-delta proof wouldn't work, since I wouldn't know the limit is equal to 2 - the question asks, if the limit exists, compute it. This was for an old analysis exam, not a calculus class, so I feel that I can't just set the above = some number L, and then make algebraic manipulations on both sides of the equation, until I get what I want. We can't assume the limit exists, I think.
Thanks,
If $$A=\sqrt{2 + \sqrt{2+\sqrt{2+\sqrt2+\cdots}}}$$ just square and get $$A^2=2+\sqrt{2 + \sqrt{2+\sqrt{2+\sqrt2+\cdots}}}=2+A$$ So $A^2-A-2=0$ and then the solution.
More generaly, if $$A(c)=\sqrt{c + \sqrt{c+\sqrt{c+\sqrt c+\cdots}}}$$ $$A^2(c)=c+A(c)$$ $$A(c)=\frac{1}{2} \left(1+\sqrt{4 c+1}\right)$$
You will find whole numbers for $c=2,6,12,20,30,42,56,72,90,110,132,156,182,\cdots$ that is to say for $c=n(n+1)$ which gives $A\big((n(n+1)\big)=n+1$.