Is $f(x)=\frac{1}{x+1} \cos x^2$ uniformly continuous?

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.

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 $[0, \infty)$.