Is $f_{n}\left(x\right)=\sin\left(x+\frac{x^{2}}{n}\right)$ uniformly convergent on $\left[0,\:2\pi\right]$

Question

Is $f_{n}\left(x\right)$ uniformly convergent on $\left[0,\:2\pi\right]$? \begin{equation} f_{n}\left(x\right)=\sin\left(x+\frac{x^{2}}{n}\right) \end{equation} We can see that $f_{n}\left(x\right)$ converges to $f\left(x\right)=\sin\left(x\right)$ point-wise.

Then how to proceed? \begin{equation} \lim_{n\rightarrow\infty}\sup_{x\in\left[0,\:2\pi\right]}\left|\sin\left(x+\frac{x^{2}}{n}\right)-\sin\left(x\right)\right| \end{equation}

And hint? or there is another way without using $\lim\sup$ ? Thank you very much.

Answer 1

Use that

$$\sin\left(x+\frac{x^{2}}{n}\right)=\sin x\cos \dfrac{x^2}{n}+\cos x\sin \dfrac{x^2}{n}.$$

Thus

\begin{align}\left|\sin\left(x+\frac{x^{2}}{n}\right)-\sin x\right|&= \left|sin x\cos \dfrac{x^2}{n}+\cos x\sin \dfrac{x^2}{n}-\sin x\right| \\ &\le |\sin x|\left|1-\cos\dfrac{x^2}{n}\right|+|\cos x|\left|\sin\frac{x^2}{n}\right|.\end{align}

Now using that

$$|\sin t|\le t, |1-\cos t|\le t, \forall t\in [0,\infty),$$

we get

$$\left|\sin\left(x+\frac{x^{2}}{n}\right)-\sin x\right|\le \dfrac{x^3}{n}+\dfrac{x^2}{n}=\dfrac{4\pi^2(2\pi+1)}{n}, \forall x\in[0,2\pi],\forall n\in \mathbb{N}.$$

So, given $\epsilon>0$ there exists $N\ge \dfrac{4\pi^2(2\pi+1)}{n}$ such that

$$n\ge N\implies\left|\sin\left(x+\frac{x^{2}}{n}\right)-\sin x\right|\le\epsilon, \forall x\in[0,2\pi],\forall n\in \mathbb{N}.$$

Answer 2

Use the inequality $$\left\vert \sin\left(x+\dfrac{x^2}{n}\right) -\sin x \right\vert \le \dfrac{x^2}{n}\le \dfrac{4\pi^2}{n}$$ and the RHS converges to zero with $n \to \infty$ independently of $x$.