Given $h_n(x) = \sqrt{x^2+\frac{1}{n}}$, I'm asked to first compute the pointwise limit of $h_n$ and then prove that it converges uniformly on $\mathbb{R}$. For pointwise convergence, I think that I'm supposed to fix $x \in \mathbb{R}$ and consider the $lim_{n\rightarrow \infty} \ h_n(x) = lim_{n\rightarrow \infty} \sqrt{x^2+\frac{1}{n}} = lim_{n\rightarrow \infty} \sqrt{x^2} = |x|$, so I say that $h_n \rightarrow h$ pointwise where $h:\mathbb{R} \rightarrow\mathbb{R}$, $h(x) = |x|$.
I'm having trouble showing that $h_n\rightarrow h$ uniformly. Given the definition of uniform convergence $\forall \epsilon > 0, \exists N(\epsilon)\in\mathbb{N}; |h_n(x)-h(x)|<\epsilon$ it seems that I would need to produce an $N(\epsilon)$ such that this inequality holds. However, I'm not having any luck with manipulating this inequality in order to find an appropriate $N(\epsilon)$. Is there another way to do it or a trick I'm not seeing? Thanks for the help!
Notice that $h_n(x)^2 - h(x)^2 = 1/n$, so: $$h_n(x) - h(x) = \frac{1/n}{h_n(x) + h(x)}$$
We have $h_n(x) + h(x) \ge \sqrt{1/n}$. Hence:
$$h_n(x) - h(x) \le \frac1{\sqrt n}$$
and this is for all $x$. Conclude.