Question about $\lim_{x \to -\infty}\frac{\sqrt{10+11x^2}}{12+13x}$ 
$\lim_{x \to -\infty}\dfrac{\sqrt{10+11x^2}}{12+13x}$

= multiply top and bottom by $\dfrac{1}{x}=-\dfrac{1}{\sqrt{x^2}}$  
My question is, why is the negative sign in front so crucial, I don't get it:  
$\lim_{x \to -\infty}-\dfrac{\sqrt{10/x^2+11x^2/x^2}}{12/x+13x/x}=-\dfrac{\sqrt{11}}{13}$  
Why wouldn't the answer be $\dfrac{\sqrt{11}}{13},$ the limit as $x$ goes to positive infinity, $x \rightarrow \infty$?  
Thank you.

Maybe this is easier to visualize:
\begin{align*}
\lim_{x \to -\infty} \frac{\sqrt{10+11x^2}}{12+13x} & = \lim_{x \to -\infty} \frac{\sqrt{x^2\big(\frac{10}{x^2}+11\big)}}{x\big(\frac{12}{x}+13\big)} \\
& = \lim_{x \to -\infty} \frac{\sqrt{x^2} \sqrt{\frac{10}{x^2}+11}}{x\big(\frac{12}{x}+13\big)} \\
& = \lim_{x \to -\infty} \frac{|x| \sqrt{\frac{10}{x^2}+11}}{x\big(\frac{12}{x}+13\big)} \\
& = \lim_{x \to -\infty} -\frac{\sqrt{\frac{10}{x^2}+11}}{\frac{12}{x}+13} \quad\text{because $\frac{|x|}{x}=-1$ when $x < 0$}\\ & =-\frac{\sqrt{11}}{13}
\end{align*}
But intuitively, if the limit exists, then it must be negative because the numerator is always positive while the denominator is negative whenever $x < -1$.
