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 $
 A: 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.
A: We have
$$\frac {\sqrt{x^2+4}+\sqrt{x^2+2}} {\sqrt{x^2+4}+\sqrt{x^2+2}}=1$$
so
$$x(\sqrt{x^2+4}-\sqrt{x^2+2})\\=x(\sqrt{x^2+4}-\sqrt{x^2+2})\cdot \frac {\sqrt{x^2+4}+\sqrt{x^2+2}} {\sqrt{x^2+4}+\sqrt{x^2+2}} \\
=\frac {x((\sqrt{x^2+4})^2-(\sqrt{x^2+2})^2)}{\sqrt{x^2+4}+\sqrt{x^2+2}}\\
=\frac {2x} {\sqrt{x^2+4}+\sqrt{x^2+2}} $$
and therefore
$$\lim x(\sqrt{x^2+4}-\sqrt{x^2+2})= \lim \frac {2x} {\sqrt{x^2+4}+\sqrt{x^2+2}}$$
if $x$ tends to an arbitrary value.
We have
$$x(\sqrt{x^2+4}-\sqrt{x^2+2})\\
=x^3 (\sqrt{1+\frac 4 {x^2} }-\sqrt{1+\frac 2 {x^2}})$$
and therefore
$$\lim x(\sqrt{x^2+4}-\sqrt{x^2+2})\\
=\lim x^3 (\sqrt{1+\frac 4 {x^2} }-\sqrt{1+\frac 2 {x^2}})$$
but we  have
$$x(\sqrt{x^2+4}-\sqrt{x^2+2})
\ne 1 (\sqrt{1+\frac 4 {x^2} }-\sqrt{1+\frac 2 {x^2}})$$
except for $x=1$, so there is no reason to assume
that the limits of the RHS and LHS of this inequality are equal except if $x\to 1$
A different example is
$$\lim_{x\to \infty}\frac {\sqrt{x^2+4}-\sqrt{x^2+2}} x $$
here you have
$$=\lim_{x\to\infty}\frac{x \sqrt{1+\frac 4 {x^2} }-\sqrt{1+\frac 2 {x^2}}}   x\\
=\lim_{x\to \infty}(\sqrt{1+\frac 4 {x^2} }-\sqrt{1+\frac 2 {x^2}})\\
=\sqrt{1+\lim_{x\to \infty} \frac 4 {x^2}}  - \sqrt{1+\lim_{x\to \infty} 
 \frac 2 {x^2}} =0$$
But you also can use the conjugate trick from the first example:
$$\frac {\sqrt{x^2+4}-\sqrt{x^2+2}} x \\
=\frac {\sqrt{x^2+4}-\sqrt{x^2+2}} x \frac {\sqrt{x^2+4}+\sqrt{x^2+2}} {\sqrt{x^2+4}+\sqrt{x^2+2}} \\
=  \frac {(\sqrt{x^2+4})^2-(\sqrt{x^2+2})^2} {x (\sqrt{x^2+4}+\sqrt{x^2+2})}\\
=\frac {2} {x (\sqrt{x^2+4}+\sqrt{x^2+2})} \in [\frac 2 {x \cdot 2(x+1)},\frac 2 {x \cdot 2x}] {\to} [0,0] \quad ({x\to \infty})
$$
