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So I've got this thing

$\lim_{x \rightarrow 3} \frac{\sqrt[3]{x^2+6x}-3}{x-3}$

I tried rationalizing but then it ended up in a huge mess and I didn't get the correct answer. I don't know if there's something else to do before resorting to rationalization. I just need something to get started.

Aside from that I know that something on the numerator will cancel out because if I evaluate the limit it ends up as $\frac{0}{0}$

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  • $\begingroup$ Use L'Hopital's rule. $\endgroup$ – msteve Jul 29 '14 at 22:46
  • $\begingroup$ Don't know what that is yet. This is my third day in class. $\endgroup$ – Argus Jul 29 '14 at 22:48
  • $\begingroup$ I would try rationalizing, as you mentioned. Multiply by $(x^2+6x)^{2/3}+3(x^2+6x)^{1/3}+9$ on the top and bottom. $\endgroup$ – user84413 Jul 29 '14 at 22:51
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Hint

$$\lim_{x \to 3} \frac{\sqrt[3]{x^2+6x}-3}{x-3}=\lim_{x\to 3} \frac{\left(\sqrt[3]{x^2+6x}-3\right)\left(\sqrt[3]{(x^2+6x)^2}+3\sqrt[3]{x^2+6x}+9\right)}{\left(x-3\right)\left(\sqrt[3]{(x^2+6x)^2}+3\sqrt[3]{x^2+6x}+9\right)}$$

$$=\lim_{x\to 3} \frac{x^2+6x-27}{\left(x-3\right)\left(\sqrt[3]{(x^2+6x)^2}+3\sqrt[3]{x^2+6x}+9\right)}=\lim_{x\to 3} \frac{x+9}{\sqrt[3]{(x^2+6x)^2}+3\sqrt[3]{x^2+6x}+9}$$

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  • $\begingroup$ I see, this is what I was looking for. Thanks for the help. $\endgroup$ – Argus Jul 29 '14 at 22:51
  • $\begingroup$ @Element118 Thank you for noticing the typo. I have fixed it. $\endgroup$ – mfl Jan 8 '16 at 20:11

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