Given $h_n(x) = (x^2 + 1/n)^{1/2}$ I am trying to prove that $h_n(x)$ converges uniformly to $h(x)$ which I calculated to be $(x^2)^{1/2}$. I understand that I need to show that $|h_n(x) - h(x)|$ can be made arbitrarily small by choosing some $N$ that does not have anything to do with $x$.
Here is my progress so far:
$$ |h_n(x) - h(x)| = |(x^2 +1/n)^{1/2} - (x^2)^{1/2}|$$ $$= |((nx^2 + 1)^{1/2})/n - (x^2)^{1/2}|$$ $$ \leq |((nx^2 +1)^{1/2})/n| $$ $$\leq |(nx^2 +1)^{1/2}|$$
After that I'm stuck, and if the problem didn't tell me to prove uniform convergence, I would probably conclude that it doesn't uniformly converge.
\begin{align} \left(x^2+\frac1n \right)^\frac12 - \left(x^2 \right)^\frac12&= \frac{\frac1n}{\left(x^2+\frac1n \right)^\frac12 + \left(x^2 \right)^\frac12} \\ &\le \frac{\frac1n}{\left(x^2+\frac1n \right)^\frac12 } \\ &\le \frac{\frac1n}{\left(\frac1n \right)^\frac12 } \\ &=\frac1{\sqrt{n}} \end{align}