# Thread: "very difficult" limits question

1. Originally Posted by topsquark
Check that again:
$\lim_{x \to \infty} \dfrac{5x^2}{\sqrt[3]{{\left( {x^3 + 5x^2 } \right)^2 }} + x\sqrt[3]{{x^3 + 5x^2 }} + x^2 }$

$= \lim_{x \to \infty} \dfrac{5}{\frac{1}{x^2}\sqrt[3]{{\left( {x^3 + 5x^2 } \right)^2 }} + \frac{1}{x}\sqrt[3]{{x^3 + 5x^2 }} + 1 }$

$= \lim_{x \to \infty} \dfrac{5}{\sqrt[3]{{\left( {1 + \frac{5}{x} } \right)^2 }} + \sqrt[3]{{1 + \frac{5}{x} }} + 1 }$

$= \frac{5}{1 + 1 + 1} = \frac{5}{3}$

-Dan
thank you

2. Originally Posted by polymerase
....wow how do you come up with something like this to solve this question? i have never seen anyone rationalize like this before
It's a well known trick.

It's based using the identity $a^3-b^3=(a-b)(a^2+ab+b^2)$

3. Originally Posted by Krizalid
It's a well known trick.

It's based using the identity $a^3-b^3=(a-b)(a^2+ab+b^2)$
"Well known" he calls it!

I had to "invent" the method for myself once when I was teaching a College Algebra course.

They don't teach us poor Physicists anything!

-Dan

4. ¿?

What ¿?

Sorry, but I thought it was well known xD!

5. Originally Posted by Krizalid
It's a well known trick.

It's based using the identity $a^3-b^3=(a-b)(a^2+ab+b^2)$
hehe, i laughed at that too, topsquark

you seem to have tweeked the eqution though. because here, "b^2" should have been "x^{2/3}" i think, you have just "x^2" ...but anyway, i kind of see where you're coming from

Originally Posted by Krizalid
¿?

What ¿?

Sorry, but I thought it was well known xD!
perhaps for you

Originally Posted by Krizalid
You're wrong
haha, and he said that with a smile. that was great!

you are the man, Krizalid!!!

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