Limit $\lim_{x\to\infty} x(\sqrt{x^2 +4} - \sqrt{x^2 + 2})$

Question

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 $

Answer 1

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:

$$\frac {\sqrt{x^2+4}-\sqrt{x^2+2}} x \\ =\frac {\sqrt{x^2+4}-\sqrt{x^2+2}} x \frac {\sqrt{x^2+4}+\sqrt{x^2+2}} {\sqrt{x^2+4}+\sqrt{x^2+2}} \\ = \frac {(\sqrt{x^2+4})^2-(\sqrt{x^2+2})^2} {x (\sqrt{x^2+4}+\sqrt{x^2+2})}\\ =\frac {2} {x (\sqrt{x^2+4}+\sqrt{x^2+2})} \in [\frac 2 {x \cdot 2(x+1)},\frac 2 {x \cdot 2x}] {\to} [0,0] \quad ({x\to \infty}) $$

Answer 2

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.