Find the limit of the following sequence (I'll post it below)

Question

anonymous · Answer

$$\sqrt{2}, \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}, ...$$ So in recursive form it would look like this:
$$a_n=\sqrt{2(a_n-1)}$$
I know that the answer to this is supposed to be 2, but I'm not sure how to get there. Any help would be appreciated.

anonymous · Answer

put $x=\sqrt{2\sqrt{2\sqrt2}..}$

anonymous · Answer

then square it

anonymous · Answer

you get $x^2=2x$

experimentx · Answer

my intuition says that you need to show that the limit exists prior to calculating it.

anonymous · Answer

@satellite73 Thanks so much, I don't know why I didn't think of that. We're just starting out with sequences and series and for some reason I find them a little tricky to work with.

@experimentX I was told that the limit existed so I didn't really need to determine that, but for next time it would probably help to know how to do that. Do you know how to show that this limit exists?

anonymous · Answer

@satellite73 Doesn't x=0 also fulfill that equation? How do I know the limit isn't 0?

experimentx · Answer

show the sequence is monotonically increasing and is bounded above.