<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] You certainly  forgot an absolute value in your computations.
Here is a computation for $x<0$:
\begin{align} 
f(x)&= \frac{x^2-\sqrt{x^4+1}}{x^3-\sqrt{x^6+1}}=-\frac 1{\Bigl(x^3-\sqrt{x^6+1}\Bigr)\Bigl(x^2+\sqrt{x^4+1}\Bigr)}\\
&= -\frac 1{\biggl(x^3-|x^3|\sqrt{1+\dfrac1{x^6}}\mkern2mu\biggr)\Bigl(x^2+\sqrt{x^4+1}\Bigr)} 
\sim_{-\infty} \frac{-1}{2x^3\cdot2x^2}=-\frac1{4x^5}\to 0^+.
\end{align}
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