calculating a limit of a function I am having hard time to proving the limit of $\lim \frac{1}{x^3}$ as $x\rightarrow +\infty $ and $x\rightarrow -\infty $.
I am pretty sure I need to use the lemma of:
if $f(x)=\frac{1}{g(x)}$ and $\lim g(x)=\pm \infty$ then $f(x)\rightarrow0$.
Does it a good way to prove it? or may I use another way? Thanks in advance.
 A: $$\lim_{x \to \infty} \frac{1}{x^3}=\frac {1}{\displaystyle \lim_{x \to \infty} x^3} =\frac {1}{(\displaystyle \lim_{x \to \infty} x)^3}=0$$
Similarly $$\lim_{x \to -\infty} \frac{1}{x^3}=\frac {1}{\displaystyle \lim_{x \to -\infty} x^3} =\frac {1}{(\displaystyle \lim_{x \to -\infty} x)^3}=0$$
A: Suppose that for all $x\gt 0$ and some $p\gt0$ that $$1/x^3\gt p^3\implies 1/x>p\implies x\lt 1/p$$
but this contradicts our assumption that $1/x^3>p^3$ for ALL $x\gt 0$, hence the lower bound for $1/x^3$ is not greater than 0 as x becomes arbitrarily large. And, because $x$ is never negative we can conclude that
$$\lim_{x\to+\infty}1/x^3=0$$
And without a loss of generality, we can also conclude that
$$\lim_{x\to-\infty}1/x^3=0$$
A: First note that
$$
\lim_{x\to \infty} f(x)=L
$$
If for every $\epsilon\gt 0$ there exists some $M\gt 0$, such that $|f(x)-L|\lt\epsilon$ whenever $x\gt M$.
So now using this definition, let's prove that
$$ \lim_{x\to \infty} \frac{1}{x^3}=0 $$
In this case, we have
$$ |f(x)-L|=\left|\frac{1}{x^3}-0\right|= \left|\frac{1}{x^3}\right|=\frac{1}{|x^3|} \lt \epsilon $$
$$ \frac{1}{\epsilon}\lt |x^3| $$
$$ |x|\gt\frac{1}{\sqrt[3]{\epsilon}}=M\gt 0 $$
Thus
$$ \lim_{x\to \infty} \frac{1}{x^3}=0 $$
For limits approaching $-\infty$, the definition is the same except that $x\lt M$ and $M\lt 0$. Give it a shot.
