More Limits !

I have to find the limit as x tends to infinity for :

x²(x³+1)^1/3-x^3

I had the idea of expanding the part in bold using a taylor series for x = infinity.

However, I have no idea how to go about doing this.

Also, wolfram alpha turns up this neat series expansion for the entire equation :



Setting x=infinity would give me a limit of 1/3.

How did they achieve this ?

Note that $(x^3+1)^{1/3} = (x^3)^{\frac{1}{3}}(1+\frac{1}{x^3})^{\frac{1}{3}}$

I have to find the limit as x tends to infinity for :

x²(x³+1)^1/3-x^3

I had the idea of expanding the part in bold using a taylor series for x = infinity.

However, I have no idea how to go about doing this.

Also, wolfram alpha turns up this neat series expansion for the entire equation :



Setting x=infinity would give me a limit of 1/3.

How did they achieve this ?

I have to find the limit as x tends to infinity for :

x²(x³+1)^1/3-x^3

I had the idea of expanding the part in bold using a taylor series for x = infinity.

However, I have no idea how to go about doing this.

Also, wolfram alpha turns up this neat series expansion for the entire equation :



Setting x=infinity would give me a limit of 1/3.

How did they achieve this ?

$[br]\displaystyle\lim_{x\to {\infty}} (x^2(x^3+1)^{\frac{1}{3}} - x^3)[br]\displaystyle\lim_{x\to {\infty}} (x^2[x^3(1+\frac{1}{x^3})]^{\frac{1}{3}} - x^3)[br]\displaystyle\lim_{x\to {\infty}} (x^3(1+\frac{1}{x^3})^{\frac{1}{3}} - x^3)[br]\displaystyle\lim_{x\to {\infty}} (x^3[(1+\frac{1}{x^3})^{\frac{1}{3}} - 1])[br]$
Now we can use the taylor expansion.
$\displaystyle\lim_{x\to {\infty}} (x^3[(1+\frac{1}{{3x}^3})- 1])$

What! is this the maths they do at Imperial chemistry!?

Where would I go from here ? Just give me a push in the right direction...

Why ? Do you find it easy ? Yes, this is part of what we do.

It seems that Aurum has thrown up a bit more working. Factorise as he has done and then consider the binomial expansion of $(1+x^{-3})^{\frac{1}{3}}$ for $|1/x^3|<1$ i.e. for large x.

Okay, I've got it now. There's just one last thing... In aurum's working; in between step 2 and 3 the x^2 disappears.

What did he do with it ?

What happened to the x^2 ? Also, am I taylor expanding only the (1+1/x^3)^1/3 part ? ?

How did you get to the last step ?

No its hard!
I am scared of gonig to Imperial now if I get in..

Farhan has answered your second and third question.
About your first question-
$(x^2(x^3+1)^{\frac{1}{3}} - x^3)[br](x^2[x^3(1+\frac{1}{x^3})]^{\frac{1}{3}} - x^3)[br]$
Now you must know that
$[x^3(1+\frac{1}{x^3})]^{\frac{1}{3}} = x(1+\frac{1}{x^3})^{\frac{1}{3}}$
Now I've simply multiplied x by x^2 to get x^3 as given in the next line of my previous post.