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how do I find the limit of these functions?

limit of x approaches infinity, $\displaystyle \lim_{x \to \infty}\displaystyle \frac{x^{10}-x^3}{x^9+1}$

limit of x approaches zero, $\displaystyle \lim_{x \to 0}\displaystyle \frac{x^3-1}{x^3+1}$

I'm pretty much confused. Thanks for the help!

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  • $\begingroup$ Have you tried guessing? $\endgroup$ – David H Sep 24 '13 at 10:56
  • $\begingroup$ The top goes to x and the bottom goes to 0? $\endgroup$ – user40424 Sep 24 '13 at 11:02
  • $\begingroup$ What does x approach as x approaches infinity? $\endgroup$ – David H Sep 24 '13 at 11:04
  • $\begingroup$ Infinity? that's obvious $\endgroup$ – user40424 Sep 24 '13 at 11:05
  • $\begingroup$ But the top is approaches x not infinity? $\endgroup$ – user40424 Sep 24 '13 at 11:05
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Hints: pay attention to the fact that in both cases we do the same "trick" with the highest power of $\,x\,$ in the expression...

$$\begin{align*}\bullet&\;\;\frac{x^{10}-x^3}{x^9+1}\cdot\frac{\frac1{x^{10}}}{\frac1{x^{10}}}=\frac{1-\frac1{x^7}}{\frac1x+\frac1{x^{10}}}\\{}\\ \bullet&\;\;\frac{x^3-1}{x^3+1}\cdot\frac{\frac1{x^3}}{\frac1{x^3}}=\frac{1-\frac1{x^3}}{1+\frac1{x^3}}\end{align*}$$

Remember: use arithmetic of limits and

$$\lim_{x\to\infty}\frac1{x^n}=0\;\;,\;\text{for}\;\;n\in\Bbb N$$

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  • 1
    $\begingroup$ He doesn't need to factorize for the second. The limit is approaching zero in the second one, if he just plugs in 0 that's fine. $\endgroup$ – user66733 Sep 24 '13 at 11:02
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    $\begingroup$ That's true, @some1.new4u .I didn't notice the second limit was for $\,x\to 0\;$ ...oh, well. $\endgroup$ – DonAntonio Sep 24 '13 at 11:12
  • $\begingroup$ Thanks dear Don for your nice supervisions :+) $\endgroup$ – mrs Sep 24 '13 at 12:23
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You just need to plug in $0$ for the second one, unless you're looking for an $\epsilon$-$\delta$ proof of it. $$\displaystyle \lim_{x \to 0} \frac{x^3-1}{x^3+1}=-1 $$

For the first one, you can use factorize as DonAntonio did or you can use L'hopital's rule if you know what it is.

In general, if you have a polynomial $P(x)=a_0+a_1x+\cdots + a_{n-1}x^{n-1}+a_nx^n$ and $x \to \infty$ then $P(x) \approx a_nx^n$. That means when dealing with the limits where $x \to \infty$ you can think that only the highest degree term of the polynomial matters.

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