# What does it mean by a function $f(x)=\exp(O(|x|^2))$ for $|x|$ large?

Given $f(x)$ is continuous in $(-\infty,\infty)$ and $f(x)=\exp(O(|x|^2))$ for $|x|$ large. Now I have an I expression like $$\lim_{t\rightarrow 1}\int_{-\infty}^{\infty}\exp(-z^2)[f(2xt/(1+t^2)+\sqrt{2(1-t^2)/(1+t^2)}z)-f(x)]dz.$$ Now since $f(x)=\exp(O(|x|^2))$ for $|x|$ large, for any $\epsilon >0$ and any fixed $x\in\mathbb{R}$, there exists a large enough number $L(\epsilon,x)>0$ and a small number $\delta_{1}(\epsilon)>0$ such that as $|t-1|<\delta_{1}$, $$|\int_{|z|>L}\exp(-z^2)[f(2xt/(1+t^2)+\sqrt{2(1-t^2)/(1+t^2)}z)-f(x)]dz|< \epsilon /2.$$ How to get this?

• en.wikipedia.org/wiki/Big_O_notation – reuns Jul 22 '15 at 4:27
• Are you confused by the meaning of $\exp$? or the meaning of $O$? or $|x|^2$? – user147263 Jul 22 '15 at 4:28
• Please clarify. Editing the text from our other post (posted errorneously as an answer) will help. Not sure whether it helps enough - leaving that for others to decide. – Jyrki Lahtonen Jul 22 '15 at 6:56

$f(x)=\exp(O(|x|^2))$ means that there is a $C$ such that $$\lim_{x \to \infty} \frac{f(x)}{\exp (C|x|^2)} \leq 1$$
• Yes $f(x)=\exp(O(|x|^2))$ – user256055 Jul 22 '15 at 4:42
• Indeed $\lim\limits_{x\to\infty}f(x)\leqslant g(x)$ makes one shiver in terror. – Did Jul 22 '15 at 13:01