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

Question

Consider $f:[0,\infty)\rightarrow \mathbb{R}$:$$f(x)=\frac{\cos \left(x\right)}{1+x^2}$$Is $f$ is uniformly continuous on $[0,\infty)$?

So far I tried to bound the derivative of this, but i get: $$\frac{d}{dx}\left(\frac{\cos \left(x\right)}{1+x^2}\right)=-\frac{\left(x^2+1\right)\sin \left(x\right)+2x\cos \left(x\right)}{\left(x^2+1\right)^2},$$ and I can't see how to bound that.
I thought about using with $\epsilon -\delta \:$ definiton but got stuck there too.
Any other ideas? Thanks in advance!

Answer 1

Bound it this way $$\left|\frac{\left(x^2+1\right)\sin \left(x\right)+2x\cos \left(x\right)}{\left(x^2+1\right)^2}\right|\leq\frac{(x^2+1)|\cdot|\sin(x)|+2|x|\cdot|\cos(x)|}{|x^2+1|^2}\leq\frac{x^2+1+2|x|}{(x^2+1)^2}\\=\frac{(|x|+1)^2}{(x^2+1)^2}<1$$ for $x$ large.

The uniform continuity in some interval $[0,M]$ near $0$, for $M$ large (actually $M=1$ is large enough) goes from Cantor's theorem.

Answer 2

Yes, $f$ is uniformly continuous. Since $\lim_{x\to \infty} f(x) = 0$, given $\epsilon > 0$, there exists a positive integer $N$ such that $|f(x)| < \epsilon/2$ for all $x \ge N$. Since $f$ is continuous on the closed interval $[0,N]$, it is uniformly continuous on $[0,N]$. So there exists $\delta > 0$ such that for all $x,y\in [0,N]$, $|x - y| < \delta$ implies $|f(x) - f(y)| < \epsilon/2$. Given $x,y\in [0,\infty)$ with $|x - y| < \delta$, either

1. $x,y\in [0,N]$, which implies

$$|f(x) - f(y)| < \frac{\epsilon}{2} < \epsilon.$$

2. $x\in [0,N], y\in [N,\infty)$, which implies

$$|f(x) - f(y)| \le |f(x) - f(N)| + |f(N) - f(y)| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$$

3. $x,y\in [N,\infty)$, which implies

$$|f(x) - f(y)| \le |f(x)| + |f(y)| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$$

4. $x\in [N,\infty), y\in [0,N]$ -- similar to $2$.

So for all $x,y\in [0,\infty)$, $|x - y| < \delta$ implies $|f(x) - f(y)| < \epsilon$. Since $\epsilon$ was arbitrary, $f$ is uniformly continuous on $[0, \infty)$.

Note. Following this argument, you can prove more generally that if $f :[0,\infty) \to \Bbb R$ is continuous such that $\lim_{x\to \infty} f(x) = 0$, then $f$ is uniformly continuous.