Indeterminate form uniform convergence limit I have $$u_n(x) = \sqrt{\bigg(\frac{2}{3}x + \frac{1}{\sqrt[3]{n^2}}\bigg)^3} - \frac{1}{n}$$
$$u(x) = \sqrt{\bigg(\frac{2}{3}x\bigg)^3}$$
and $u_n(x) \rightarrow u(x)$.
I have to check uniform convergence in $[0, +\infty)$:
$$\sup_{x \in [0, +\infty)}|u_n(x) - u(x)| \rightarrow  ?$$
$u_n(x)$ and $u(x)$ have positive derivatives in $[0, +\infty)$, so their sup is $+\infty$, and I have an indetermined form $+\infty -\infty$ in the limit above.
I thought that I could consider $ \forall b > 0$: $$\sup_{x \in [0, b]}|u_n(x) - u(x)| = |u_n(b) - u(b)| \rightarrow 0$$ 
and then state that we have the same result in $[0, +\infty)$. Is it correct?
Any help would be very appreciated, thanks!

Hint. By Taylor expansion, for $n\geq 1$ and for $t>0$,
$$\bigg(t + \frac{1}{n^{2/3}}\bigg)^{3/2}=t^{3/2}\bigg(1 + \frac{1}{tn^{2/3}}\bigg)^{3/2}=t^{3/2}\bigg(1 + \frac{3}{2tn^{2/3}}+\frac{3}{8t^2n^{4/3}}+o(1/t^2)\bigg).$$
So, given $n\geq 1$, what is the limit (note that $t=\frac{2x}{3}$)
$$\lim_{x\to +\infty}|u_n(x) - u(x)|=\lim_{t\to +\infty}\left|
\bigg(t + \frac{1}{n^{2/3}}\bigg)^{3/2}-\frac{1}{n}-t^{3/2}\right|?$$
What is $\sup_{x \in [0, +\infty)}|u_n(x) - u(x)|$?
