Prove that $f_n(x)=\sin{\sqrt{x+4n^2\pi ^2}}\,,\;x\geq 0$ is equicontinous on $[0,+\infty)$

Question

Prove that $f_n(x)=\sin{\sqrt{x+4n^2\pi ^2}}\,,\;x\geq 0$

1. is equicontinous on $[0,+\infty)$.

2. converges pointwise to $0$ on $[0,+\infty)$.

MY TRIAL

2. \begin{align}\sqrt{x+4n^2\pi ^2}&=\sqrt{x+4n^2\pi ^2}\left[\dfrac{\sqrt{x+4n^2\pi ^2}+2n\pi}{\sqrt{x+4n^2\pi ^2}+2n\pi}\right]\\&=\left[\dfrac{x+4n^2\pi ^2+2n\pi\sqrt{x+4n^2\pi ^2}}{\sqrt{x+4n^2\pi ^2}+2n\pi}\right]\\&=\left[\dfrac{4n^2\pi ^2+2n\pi\sqrt{x+4n^2\pi ^2}+x}{\sqrt{x+4n^2\pi ^2}+2n\pi}\right]\\&=2n\pi+\dfrac{x}{\sqrt{x+4n^2\pi ^2}+2n\pi}\end{align} As $n\to \infty,$ \begin{align}\dfrac{x}{\sqrt{x+4n^2\pi ^2}+2n\pi}\to 0\end{align} Hence, $f_n(x)=\sin{\sqrt{x+4n^2\pi ^2}}\to 0$ as $n\to 0$

1.)

Let $x,y\in [0,+\infty)$ such that $|x-y|<\delta$. We show that \begin{align}\left|f_n(x)-f_n(y)\right|<\epsilon\end{align} Now, \begin{align}\left|f_n(x)-f_n(y)\right|=\left|\sin{\sqrt{x+4n^2\pi ^2}}-\sin{\sqrt{y+4n^2\pi ^2}}\right|\end{align} I'm stuck here! Please, how should I go?

Answer 1

By mean value theorem： $$\left|f_n(x)-f_n(y)\right|=\left|\sin{\sqrt{x+4n^2\pi ^2}}-\sin{\sqrt{y+4n^2\pi ^2}}\right|= |f_n'(\xi_n)||x-y|\leq|x-y|.$$