How to find the limit of $\lim_{x\to -\infty}\frac{3x^3+6x^2+45}{5|x|^3+25|x|+12}$ 
Evaluate $$\lim_{x\to -\infty}\dfrac{3x^3+6x^2+45}{5|x|^3+25|x|+12}$$

Is just a matter dividing all variables by $x^3$ and getting $\frac{3}{5}$?
I tried looking it up and saw the graph doesn't just stop at $0$.

Since $x<0$ we get $|x|=-x$ , therefore $|x|^3=-x^3$ . Hence, the limit becomes :
 $$\
\mathop {\lim }\limits_{x \to  - \infty } \frac{{3x^3  + 6x^2  + 45}}{{ - 5x^3  - 25x + 12}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{x^3 (3 + \frac{6}{x} + \frac{{45}}{{x^3 }})}}{{x^3 ( - 5 - \frac{{25}}{{x^2 }} + \frac{{12}}{{x^3 }})}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{3x^3 }}{{ - 5x^3 }} =  - \frac{3}{5}.
\
$$
