Thread: "very difficult" limits question

1. "very difficult" limits question

$
\displaystyle\lim_{x\to\infty}\sqrt[3]{(x^3+5x^2)}-x
$

2. How about multiplyin' by

$\frac{{\sqrt[3]{{\left( {x^3 + 5x^2 } \right)^2 }} + x\sqrt[3]{{x^3 + 5x^2 }} + x^2 }}
{{\sqrt[3]{{\left( {x^3 + 5x^2 } \right)^2 }} + x\sqrt[3]{{x^3 + 5x^2 }} + x^2 }}$
?

3. Originally Posted by Krizalid
How about multiplyin' by

$\frac{{\sqrt[3]{{\left( {x^3 + 5x^2 } \right)^2 }} + x\sqrt[3]{{x^3 + 5x^2 }} + x^2 }}
{{\sqrt[3]{{\left( {x^3 + 5x^2 } \right)^2 }} + x\sqrt[3]{{x^3 + 5x^2 }} + x^2 }}$
?
huh?

4. Originally Posted by polymerase
huh?
He's showing you how to rationalize the "numerator" of your expression. (It requires more terms for a cube root than it does for a square root.) Try it and see what happens.

-Dan

5. Originally Posted by polymerase
$
\displaystyle\lim_{x\to\infty}\sqrt[3]{(x^3+5x^2)}-x
$
If you know these methods you can try:

$
\sqrt[3]{x^3+5x^2}-x=x~ \sqrt[3] {1+5/x} -x
$

Expand the cube root using the binomial theorem:

$
\sqrt[3]{x^3+5x^2}-x=x~ \sqrt[3] {1+5/x} -x=x~\left(1+\frac{5}{3x}+O(x^{-2})\right) -x=\frac{5}{3}+O(x^{-1})
$

Hence the required limit is $5/3$.

RonL

6. Originally Posted by CaptainBlack
If you know these methods you can try:

$
\sqrt[3]{x^3+5x^2}-x=x~ \sqrt[3] {1+5/x} -x
$

Expand the cube root using the binomial theorem:

$
\sqrt[3]{x^3+5x^2}-x=x~ \sqrt[3] {1+5/x} -x=x~\left(1+\frac{5}{3x}+O(x^{-2})\right) -x=\frac{5}{3}+O(x^{-1})
$

Hence the required limit is $5/3$.

RonL
how did u use the binomial theorem when its a fractional exponent?

7. Originally Posted by Krizalid
How about multiplyin' by

$\frac{{\sqrt[3]{{\left( {x^3 + 5x^2 } \right)^2 }} + x\sqrt[3]{{x^3 + 5x^2 }} + x^2 }}
{{\sqrt[3]{{\left( {x^3 + 5x^2 } \right)^2 }} + x\sqrt[3]{{x^3 + 5x^2 }} + x^2 }}$
?
I tried this way and in the end, I end up with a nice numerator but not so nice denominator:

$\dfrac{5x^2}{\sqrt[3]{{\left( {x^3 + 5x^2 } \right)^2 }} + x\sqrt[3]{{x^3 + 5x^2 }} + x^2 }$

So how would u deal with the denominator now?

8. Originally Posted by polymerase
I tried this way and in the end, I end up with a nice numerator but not so nice denominator:

$\dfrac{5x^2}{\sqrt[3]{{\left( {x^3 + 5x^2 } \right)^2 }} + x\sqrt[3]{{x^3 + 5x^2 }} + x^2 }$

So how would u deal with the denominator now?
Multiply the numerator and denominator by $\frac{1}{x^2}$, then take your limit.

-Dan

9. Originally Posted by topsquark
Multiply the numerator and denominator by $\frac{1}{x^2}$, then take your limit.

-Dan
thanks

10. Originally Posted by polymerase
how did u use the binomial theorem when its a fractional exponent?
The binomial theorem applies for any real exponent.

A MacLaurin/Taylor expansion will do just as well if you don't like a binomial expansion

RonL

11. Originally Posted by Krizalid
How about multiplyin' by

$\frac{{\sqrt[3]{{\left( {x^3 + 5x^2 } \right)^2 }} + x\sqrt[3]{{x^3 + 5x^2 }} + x^2 }}
{{\sqrt[3]{{\left( {x^3 + 5x^2 } \right)^2 }} + x\sqrt[3]{{x^3 + 5x^2 }} + x^2 }}$
?
Your method gives u this:

$
\frac{5}{1+\(1+\frac{5}{x}}
$

12. Originally Posted by Krizalid
How about multiplyin' by

$\frac{{\sqrt[3]{{\left( {x^3 + 5x^2 } \right)^2 }} + x\sqrt[3]{{x^3 + 5x^2 }} + x^2 }}
{{\sqrt[3]{{\left( {x^3 + 5x^2 } \right)^2 }} + \sqrt[3]{{x^3 + 5x^2 }} + x^2 }}$
?
Using your method gives me this:

$
\lim_{x\to\infty}\dfrac{5}{\sqrt[3]{{\left( {1 + \frac{5}{x} } \right)^2 }} + \sqrt[3]{{\frac{1}{x^3} + \frac{5}{x^4} }} + 1 }
$

and when you take the limit to infinity you get $\frac{5}{2}$ but the answer is $\frac{5}{3}$

13. You're wrong

14. Originally Posted by polymerase
Using your method gives me this:

$
\lim_{x\to\infty}\dfrac{5}{\sqrt[3]{{\left( {1 + \frac{5}{x} } \right)^2 }} + \sqrt[3]{{\frac{1}{x^3} + \frac{5}{x^4} }} + 1 }
$

and when you take the limit to infinity you get $\frac{5}{2}$ but the answer is $\frac{5}{3}$
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

15. Originally Posted by Krizalid
You're wrong
ok ok your right....my bad i forgot the x in the 2 term
Thanks man
....wow how do you come up with something like this to solve this question? i have never seen anyone rationalize like this before

Page 1 of 2 12 Last