Proving uniform convergence of $f_n(z)=\frac{4n\sqrt{nz}}{3+4n^2z}$

Question

Study the convergence and uniform convergence of the functions $$f_n(z)=\frac{4n\sqrt{nz}}{3+4n^2z},\:z\in[\delta,+\infty],\delta>0$$.

$$\lim_{n\to\infty}\frac{4n\sqrt{nz}}{3+4n^2z}=\lim_{n\to\infty}\frac{4z^{\frac{1}{2}}}{\frac{3}{n^{\frac{1}{2}}}+4n^{\frac{1}{2}}z}=0$$ So the function converges pointwise to $0$.

For uniform convergence I did the following:

$|\frac{4n\sqrt{nz}}{3+4n^2z}-0|\leqslant\frac{4n\sqrt{nz}}{4n^2z}=\frac{1}{n^{\frac{1}{2}}z^{\frac{1}{2}}}=\frac{1}{n^{\frac{1}{2}}\delta^{\frac{1}{2}}}\to 0$ as $n\to\infty$

Questions:

Is this proof right? If not why? Which are the alternatives?

Thanks in advance!

Answer 1

Your proof seems correct to me. The general idea of uniform convergence is to show that you can find some expression independent of the choice of $z$ that converges to the required limit "at least as fast" as the sequence of functions. By using the boundary of the domain of the functions (the left boundary - $\delta$), you have found a sequence independent of $z$ that converges to $0$ - and so you have uniform convergence on the entire domain.