I believe that the upper limit is $+2$ and the lower limit is $-2$. We have the trigonometric identity$$\cos x^2 -\cos (x+1)^2 =2\sin(x^2+x+\frac{1}{2})\sin(x+\frac{1}{2}).$$
We then make the substitution $x=m2\sqrt{\pi}$, where $m$ is a positive integer, so that$$f(m2\sqrt{\pi})=2\sin^2(m2\sqrt{\pi}+ \frac{1}{2}).$$
From here we use the fact that for irrational $\gamma$, the set$$S_\gamma =\{a+b\gamma\mid a\in\mathbb{Z}, b\in\mathbb{N}\}$$is dense in the reals, together with the continuity of sine to prove that $f$ can get arbitrarily close to $+2$ on any domain of the form $(x,\infty)$. We can show that the lower limit of $f$ is $-2$ using similar reasoning and the substitution $x=(2n+1)\sqrt{\pi}$.
Is this proof right? Is there another proof that doesn't use the density lemma or even better, that doesn't discretize the domain (ie. doesn't restrict the domain to a countable set)? (I can give a proof of the density lemma, but it hasn't been checked.)