Find: $\lim_{x\to -\infty} f(x)= \frac{x^2-\sqrt{x^4+1}}{x^3-\sqrt{x^6+1}}$ (no L'Hopital) 
Find: $\displaystyle \lim_{x\to -\infty} f(x)= \frac{x^2-\sqrt{x^4+1}}{x^3-\sqrt{x^6+1}}$ (no L'Hopital)

After developing the expression, by multiplying the fraction by the conjugates and rearranging, I found:
$$f(x)=x\times \frac{1+\sqrt{1+1/x^6}}{1+\sqrt{1+1/x^4}}$$
I can solve it for the limit when $x\to \infty$, with answer $\infty$ which is correct (by the book and Wolfram Alpha). But when the limit is $x\to -\infty$ the answer is zero, but from this expression, I get $-\infty$ as the answer.
I'm certainly missing something. 
Hints and answers appreciated. Sorry if this is a duplicate.

Let $y=-x:$
$F(y) = \dfrac{y^2 -\sqrt{y^4 +1}}{-y^3 - \sqrt{y^6+1}}=$
$(-1)\dfrac{y^2 - \sqrt{y^4+1}}{y^3 +\sqrt{y^6+1}}.$
Conjugates and rearrangement , and then considering $\lim_{y \rightarrow \infty}$ should get you there .
Or:
$0\le |F(y)| \le \dfrac{2\sqrt{y^4+1}}{2y^3} \lt \dfrac{4y^2}{2y^3}$.
And the limit is?.
Used:
$ |y^2 - \sqrt{y^4+1}| \lt 2\sqrt{y^4+1} \lt 4y^2.$
