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

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 $\sqrt[2]{x}$. Lets call this limit $N$

Then the largest $x$ we could have would be ${N}^{2}$, which is not even near $\infty$

(Of course there is a lower limit to both $x$ and $\sqrt[2]{x}$.
they both can't be negative)