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

Question

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?

Answer 1

$f(x)=\exp(O(|x|^2))$ means that there is a $C$ such that $$\lim_{x \to \infty} \frac{f(x)}{\exp (C|x|^2)} \leq 1$$