Limit $\lim_{x\to\infty} x(\sqrt{x^2 +4} - \sqrt{x^2 + 2})$ So i have this limit:
$$\lim_{x\to\infty} x(\sqrt{x^2 +4} - \sqrt{x^2 + 2})$$
Why can't I just divide by $\sqrt{x^2}$? Outside of the square roots you'd divide by $x$ (or $-x$). If I do this I'd get 1(1-1) which could be an answer right? The answer is $-\frac{3}{2}$ and I saw the solutions which make sense but I don't get why my method isn't right.
It has to be done without l'Hopital by the way!
EDIT sorry i wrote down the wrong answer from a different exercise
This is what I did:
$$\lim_{x\to\infty} x(\sqrt{x^2 +4} - \sqrt{x^2 + 2}) = \lim_{x\to\infty} 1(\sqrt{1 + \frac{4}{x^2}} - \sqrt{1 + \frac{2}{x^2}})$$
And since $\frac{4}{x^2}$, $\frac{2}{x^2}$ are $0$, I thought the answer would be $1(1-1) = 0 $

You can multiply and divide by the conjugate:
$$
x(\sqrt{x^2+4}-\sqrt{x^2+2})= \dfrac{2x}{\sqrt{x^2+4}+\sqrt{x^2+2}}=\dfrac{2}{\sqrt{1+4/x^2}+\sqrt{1+2/x^2}} \to 1 \quad (x \to +\infty)
$$
What you proposed, simply dividing by $\sqrt{x^2}$, leads to a different limit. You would also have to multiply by the same expression, leaving you more or less at the starting point.
