Prove by L'Hospital's rule that $\lim\limits_{x\to\infty}(\sqrt{x}(1-\delta^2)^x)=0$, $\delta\in(0,1]$ Prove the following with the help of L'Hospital's rule:
$$\lim\limits_{x\to\infty}\sqrt{x}(1-\delta^2)^x=0$$ where $\delta\in(0,1]$ is a constant.
I broke up $\sqrt{x}(1-\delta^2)^x$ into $\sqrt{x}$ and $(1-\delta^2)^x$. Computing respective derivatives yields $$\frac{d(\sqrt{x})}{dx} = \frac{1}{2\sqrt{x}}$$ and $$\frac{d((1-\delta^2)^x)}{dx} = (1-\delta^2)^x\ln(1-\delta^2)$$ and then I am stuck as to what do I do with $x$ power.
Any help is appreciated!

Since $\delta\in(0,1)$ we have $1-\delta^2<1$. Defining $\Delta:=(1-\delta^2)^{-1}>1$, L'Hopital gives $$\lim\limits_{x\to\infty}\frac{x^{1/2}}{\Delta^x}=\lim_{x\to\infty}\frac{(2x^{1/2})^{-1}}{\Delta^x\log\Delta}=\frac1{2\log\Delta}\lim_{x\to\infty}\frac1{\Delta^x\sqrt x}=0.$$
