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!

I understand the the OP requires the use of L'Hopitals rule but there is really no need for that. This is a poorly designed exercise.
$$
\lim_{x\to +\infty}\sqrt{x}(1-\delta 
^2)^x = \exp \left\{\lim_{x\to +\infty} \left(\frac12 \ln x + x \underbrace{\ln(1-\delta^2)}_{< 0}\right)\right\} = \exp \{-\infty\} = 0
$$
