# arithmetic between limits

I need find $\lim\limits_{x\rightarrow\infty}( x^3-x^2)=L$. In this case, I ended up with infinity minus infinity. But I do know $x^3$ is always greater than $x^2$ for $x\ge1$ and the difference between them is growing... Can I say $L$ is infinity? Is there a formal way to prove or show this?

Thanks!

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I would avoid arguments such as "this infinity is greater than this one..."; you have the right idea (especially since you note the difference between them is growing), but it's a bit vague and can sometimes mislead you.

To find the limit rigorously, use the following hint: Write $x^3-x^2 =x^3(1-{1\over x})$.

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what if the original expression I have is lim n!-2^n? –  cssl Feb 19 '12 at 20:08
@cssl That would be harder. What I have above works nicely for any polynomial, though. For $n!-2^n$, you might write$$n!-2^n=2^n\Bigl(\underbrace{{n\over2}\cdot{n-1\over 2}\cdot{n-2\over 2}\cdots{3\over2}\cdot{2\over2}\cdot{1\over 2}}_{\ge2}-1\Bigr)$$ for $n\ge 5$. –  David Mitra Feb 19 '12 at 20:16