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!
 A: 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 $\sup_{x \in [0, +\infty)}|u_n(x) - u(x)|$?
A: The difference of two increasing functions is not increasing in general. In your case it is since the derivative of the difference. 
$$(u_n-u)'(x)=\left(\frac23x+\frac1{n^{3/2}}\right)^{1/2}-\left(\frac23x\right)^{1/2}>0.$$
Since $u_n-u>0$ you get
\begin{align}\sup_{x\in[0,\infty)}|u_n(x)-u(x)|=\lim_{x\to\infty}(u_n(x)-u(x))=\lim_{x\to\infty}\frac32 \sqrt{\left(\frac23x+\frac1{n^{3/2}}\right)^{3}}-\frac32 \sqrt{\left(\frac23x\right)^{3}}-\frac1n
\\=\lim_{x\to\infty}\frac32\frac{\left(\frac23x+\frac1{n^{3/2}}\right)^{3}-\left(\frac23x\right)^{3}}{ \sqrt{\left(\frac23x+\frac1{n^{3/2}}\right)^{3}}+ \sqrt{\left(\frac23x\right)^{3}}}-\frac1n\end{align}
where we used $a^2-b^2=(a-b)(a+b)$. Now develop the cube and do some manipulations. After cancellations the numerator goes like $x^2$ and the denominator like $x^{3/2}$ so the limit should be...
