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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 $0$.

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  • $\begingroup$ Yeah that's more or less all there is to it, but be careful with signs!! When $x<0$, $x^3$ is negative, but $\vert x\vert^3$ is positive. $\endgroup$
    – Mathmo123
    Sep 12, 2014 at 1:40
  • $\begingroup$ Absolute values are a pain! You should always simplify to get rid of them if you can. (OK. . . maybe not always.) $\endgroup$
    – David
    Sep 12, 2014 at 1:42

3 Answers 3

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Hint: If $x<0$ then $$ |x^{3}|=-x^{3} $$

$$ |x|=-x $$

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

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Hint, substitute $-x$ for all the $x$ in the limit.

Now instead as $x\to-\infty$, $-x\to\infty$

Hopefully this will help you see the limit more clearly.

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