I want to find the $$\lim_{x\rightarrow \infty} (\sqrt x-x)$$
What I was thinking was to multiply by the conjugate since that's normally what you do with square roots, So I would get
$$\lim_{x\rightarrow\infty} (\sqrt{x}-x)\cdot\frac{(\sqrt{x}+x)}{(\sqrt{x}+x)}$$
$=\lim_{x\rightarrow\infty} \frac{x-x^2}{\sqrt{x}+x} = \lim_{x\rightarrow\infty}\frac{x(1-x)}{\sqrt{x} (1+x)} = \lim_{x\rightarrow\infty} \frac{-x}{\sqrt{x}}$
I guess as x goes to infinity, $\sqrt \infty$ is smaller than $-\infty$, so it blows up to $-\infty$ but I'm unsure if this logic is correct.
One solution I found was $\lim_{x\rightarrow\infty} \sqrt x (1-\sqrt x) = (\infty) (1-\infty) = (\infty)(-\infty) = -\infty$
However I'm just uncomfortable actually plugging in $\infty$ for limit values and I also wouldn't think to do this. I would think to multiply by the conjugate