Question

find the limit as x approaches infinity of (x^6+18x^4)^(1/3)-x^2

Answer 1

The correct answer is 6. I just don't know how to get that.

Answer 2

Factor \(x^6\) out of the radicand,\[\lim_{x \rightarrow \infty} \sqrt[3]{x^6\left( 1+\frac{ 18 }{ x^2 } \right)} - x^2\]

Answer 3

Then I get \[\lim_{x \rightarrow \infty} x^2\sqrt[3]{1+18/x^2}-x^2\]

Answer 4

OK What happens when you factor out the common factor of x^2? (Hope I'm not leading you down the garden path)

Answer 5

You're right, the correct answer is 6. Perhaps my thoughts are leading nowhere.

Answer 6

That would give me \[x^2(\sqrt[3]{\frac{ 18 }{ x^2 }+1}-1)\]

Answer 7

which means as x goes to infinity, the limit will also be going to infinity, which is not right

Answer 8

Right-o. I was looking at the quantity in brackets going to zero.

Answer 9

Oh, never mind. So it's infinity times zero.

Answer 10

But how do you continue the problem?

Answer 11

Thinking...

Answer 12

A series expansion of that expression looks like\[6-\frac{ 36 }{ x^2 }+\frac{ 360 }{ x^4 }-\frac{ 4320 }{ x^6 }+...\]which shows how the limit approaches 6.
But there must be an easier way.

Answer 13

Wait, how did you do that?

Answer 14

It's a Laurent series expansion. I looked it up. But it's a long time since I've done them. You'd be better off researching doing Laurent expansions. I'd just mess it up trying to remember how to do it.

Answer 15

I did it! Thanks for starting me off, it actually helped!!!

Answer 16

Good for you! Job well done