What is the limit as x approaches infinity of #sqrt(x)#?

1 Answer
Jan 12, 2015

There is no upper limit.

You can prove this by considering that:

If #x# goes up you always need a greater number, that, if squared, will give you #x#.

Or, the other way around:
Suppose there is a limit to #root 2 x#. Lets call this limit #N#

Then the largest #x# we could have would be #N^2#, which is not even near #oo#

(Of course there is a lower limit to both #x# and #root 2 x#.
they both can't be negative)