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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?

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  • $\begingroup$ en.wikipedia.org/wiki/Big_O_notation $\endgroup$ – reuns Jul 22 '15 at 4:27
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    $\begingroup$ Are you confused by the meaning of $\exp$? or the meaning of $O$? or $|x|^2$? $\endgroup$ – user147263 Jul 22 '15 at 4:28
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    $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Jul 22 '15 at 6:56
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$f(x)=\exp(O(|x|^2))$ means that there is a $C$ such that $$\lim_{x \to \infty} \frac{f(x)}{\exp (C|x|^2)} \leq 1$$

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    $\begingroup$ If you don't know what the question asks... what are you answering? $\endgroup$ – user147263 Jul 22 '15 at 4:37
  • $\begingroup$ Yes $f(x)=\exp(O(|x|^2))$ $\endgroup$ – user256055 Jul 22 '15 at 4:42
  • $\begingroup$ Indeed $\lim\limits_{x\to\infty}f(x)\leqslant g(x)$ makes one shiver in terror. $\endgroup$ – Did Jul 22 '15 at 13:01

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