<s>[INST] 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.
 [/INST] For a “direct” proof one can proceed as follows. Without loss of generality we can assume that $y \le x$. Then
$$
\frac{\cos x^2}{x+1}-\frac{\cos y^2}{y+1} = \frac{(y+1)\cos x^2-(x+1)\cos y^2}{(x+1)(y+1)} \\
= \frac{\cos x^2 - \cos y^2}{x+1} - \frac{(x-y)\cos y^2}{(x+1)(y+1)}
$$
(The idea here is to split the difference into one term containing the difference $\cos x^2 - \cos y^2$, and another term containing the difference $x-y$.)
The second term is easy to estimate:
$$
\left|\frac{(x-y)\cos y^2}{(x+1)(y+1)} \right| \le |x-y| \, .
$$
For the first term we need the trigonometric identity
$$
 \cos a - \cos b = -2 \sin \frac{a+b}{2} \sin \frac{a-b}{2} 
$$
and get
$$
\left | \frac{\cos x^2 - \cos y^2}{x+1} \right | =
\left | \frac{2 \sin \frac{x^2+y^2}{2}\sin \frac{x^2-y^2}{2}}{x+1}\right | \\
\le \frac{2 |\sin \frac{x^2-y^2}{2}|}{x+1} \le \frac{|x^2-y^2|}{x+1}
= \frac{x+y}{x+1} |x-y| \le 2|x-y| \, .
$$ 
Combining these estimates we get
$$
|f(x) - f(x) | \le 3 |x-y| 
$$
so that $f$ is Lipschitz continuous (and therefore uniformly continuous) on $[0, \infty)$.
</s>