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I was wondering if anyone could maybe help me start off an idea on how to prove, respectively, that $\lim \limits_{x\to\pm\infty} f(x) = \pm \infty.$

Also, $\lim\limits_{x\to 0} f(x) ,\quad \lim\limits_{ x\to 0-} f(x),\quad \lim\limits_{x\to0+} f(x)$.

where $\displaystyle f(x) = \frac{x^3} {\lvert x \rvert} $.

I know I need to use epsilon-delta but not sure how to apply?

Thank you!

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    Please, try to make the title of your question more informative. E.g., *Why does a imply a+c?* is much more useful for other users than *A question about inequality.* From [How can I ask a good question?](http://meta.math.stackexchange.com/a/589/): *Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader.*2012-11-12

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The standard $\epsilon-\delta$ definitions do not apply to limits with infinity because $\infty\pm\epsilon$ has no meaning.

Instead, the definition becomes something like this for real valued functions:

$\lim_{x\rightarrow c}f(x)=\infty$ means that for all $n\in\mathbb{N}$, there exists $\delta>0$ such that for all $x$, if $|x-c|<\delta$, then $f(x)>n$.

This encapsulates what it means for the function to "get close to infinity."

As an exercise, you can write what it means for $f(x)$ to have limit $-\infty$ as $x$ approaches $c$.


For your $x^3/|x|$ problem, looking at the left and right limits helps you get rid of the absolute value signs. If $x\rightarrow 0^-$, then $x$ is a small negative number approaching zero, and your function is $x^3/|x|=x^3/(-x)=-x^2$. On the other hand, for $x\rightarrow 0^+$, it is a shrinking positive number, so $x^3/|x|=x^3/x=x^2$.

The two-sided limit exists iff you find the left and the right limits exist, and they both are the same number.

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    okay thank you i'll try to apply that to some similar questions as well2012-11-12