<s>[INST] Is $f_{n}\left(x\right)=\sin\left(x+\frac{x^{2}}{n}\right)$ uniformly convergent on $\left[0,\:2\pi\right]$ 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.
 [/INST] 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
$$n\ge N\implies\left|\sin\left(x+\frac{x^{2}}{n}\right)-\sin x\right|\le\epsilon, \forall x\in[0,2\pi],\forall n\in \mathbb{N}.$$
</s>