<s>[INST] 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.
 [/INST] Set $x=-y\to+\infty$
$$\frac{x^2-\sqrt{x^4+1}}{x^3-\sqrt{x^6+1}}=\frac{y^2-\sqrt{y^4+1}}{-y^3-\sqrt{y^6+1}}=-\frac{y^2-\sqrt{y^4+1}}{y^3+\sqrt{y^6+1}}\frac{y^2+\sqrt{y^4+1}}{{y^2+\sqrt{y^4+1}}}=\frac{1}{(y^3+\sqrt{y^6+1})({y^2+\sqrt{y^4+1})}}\to0$$
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