Is $f(x)=\frac{1}{x+1} \cos x^2$ uniformly continuous? 
Let $f:[0,\infty)\to \Bbb{R}$, $f(x)=\frac{1}{x+1} \cos x^2$, Is $f$ uniformly continuous?

My attempt:
Let $x,y\in [0,\infty),$ then
\begin{align*}
\Bigg|\frac{\cos x^2}{x+1}-\frac{\cos y^2}{y+1}\Bigg|
& =\Bigg|\frac{(y+1)\cos x^2-(x+1)\cos y^2}{(x+1)(y+1)}\Bigg| \\
& =\Bigg|\frac{(y \cos x^2 - x \cos y^2)+(\cos x^2 - \cos y^2)}{(x+1)(y+1)}\Bigg| \\
& \le \Bigg|\frac{y \cos x^2 - x \cos y^2}{(x+1)(y+1)}|+|\frac{\cos x^2 - \cos y^2}{(x+1)(y+1)}\Bigg| \\
& \le \Big|y \cos x^2 - x \cos y^2| + |\cos x^2 - \cos y^2\Big| \\
\end{align*}
How can I complete, please? Thanks.

Hint
One can prove that if $x\mapsto f(x)$ is continuous on $[0,\infty )$ and $\lim\limits_{x\to \infty }f(x)$ exist, then $f$ is uniformly continuous on $[0, \infty )$. The proof goes as follow : 
Let $\varepsilon >0$. Since $\lim\limits_{x\to \infty }f(x)=\ell$, there is $M$ s.t. $|f(x)-\ell|<\frac{\varepsilon}{2} $ if $x,y\geq M$. In particular, if $x,y\geq M$, then $$|f(x)-f(y)|\leq |f(x)-\ell|+|f(y)-\ell|\leq \varepsilon .$$
The fact that $f$ uniformly continuous on $[0,M]$ is a famous theorem. 
Look what happen when $x\leq M\leq y$, and conclude. 
