Find constant for upper bound

Question

I want to show that there is a constant $C_{\epsilon}$ such that $$C\left(\frac{\sqrt{t}+x}{\sqrt{t}}\right)^{\alpha}\leq C_{\epsilon}\exp\left(\epsilon\frac{x^{2}}{t}\right)$$ for every $\epsilon>0$ where everything is positive here and $C$ and $\alpha$ are given constants.
Of course one can write the left hand side as: $$\exp(\ln(C))\exp(\alpha \ln \left(1+\frac{x}{\sqrt{t}}\right)).$$ Then I am stuck. Can someone help?

Answer 1

$Ce^{-\epsilon y^{2}} (1+y)^{\alpha}$ is a continuous function of $y$ and it tends to $0$ as $ y \to \infty$. It tends to $C$ as $ y \to 0$. Hence it is a bounded function on $(0,\infty)$. Let $C_{\epsilon}$ be a bound for this function. Then $Ce^{-\epsilon y^{2}} (1+y)^{\alpha} \leq C_{\epsilon}$ and putting $y=\frac x {\sqrt t}$ finishes the proof.