(x^3+x^2)^(1/3)-x

[darklight]

Hey, i need a bit of help.
Lim as x approaches infinity of the above function.
a^3-b^3=(a-b)(a^2+ab+b^2) was given as a hint.

 

[Quaid]

ֺֺ

(x^3 + x^2)^(1/3) - x

Lim as x approaches infinity

I'm having a mental block, regarding the hint, but I can calculate this limit using l'Hôpital's rule (after algebraic rearrangement into the 0/0 indeterminant form shown below).

\(\displaystyle \dfrac{\left(1 + \dfrac{1}{x}\right)^{1/3} - 1}{\dfrac{1}{x}}\)

Would you like to try this approach?

ֺֺ

 

[pka]

Hey, i need a bit of help.
Lim as x approaches infinity of the above function.
a^3-b^3=(a-b)(a^2+ab+b^2) was given as a hint.

Take note that \(\displaystyle \displaystyle\left( {\sqrt[3]{a} - {b}} \right) = \left( {\sqrt[3]{a} - {b}} \right)\left( {\frac{{\sqrt[3]{{{a^2}}} + \sqrt[3]{a}{b} + {{{b^2}}}}}{{\sqrt[3]{{{a^2}}} + \sqrt[3]{a}{b} + {{{b^2}}}}}} \right) = \left( {\frac{{a - b^3}}{{\sqrt[3]{{{a^2}}} + \sqrt[3]{a}{b} + {{{b^2}}}}}} \right)\)

Let \(\displaystyle a = {x^3} + {x^2}\,\& \,b = x\)

 

[darklight]

ֺֺ


I'm having a mental block, regarding the hint, but I can calculate this limit using l'Hôpital's rule (after algebraic rearrangement into the 0/0 indeterminant form shown below).

\(\displaystyle \dfrac{\left(1 + \dfrac{1}{x}\right)^{1/3} - 1}{\dfrac{1}{x}}\)

Would you like to try this approach?

ֺֺ

Using L'hopital's rule from here, wouldn't you be dividing by 0?

 

[ahorn]

Using L'hopital's rule from here, wouldn't you be dividing by 0?

When I applied l'Hospital's rule, I got

\(\displaystyle \lim_{x \to \infty} \frac{\frac{1}{3}\left(1+\frac{1}{x}\right)^{-2/3}\cdot-x^{-2}}{-x^{-2}}\) from the derivatives


and then \(\displaystyle \lim_{x \to \infty} \left[\frac{1}{3}(1+\frac{1}{x})^{-2/3}\right]\) when I divided through by \(\displaystyle -x^{-2}\),


which equals \(\displaystyle \frac{1}{3}\).



I didn't know how to use the difference of cubes method

 

[Quaid]

I got lim_x-> inf (1/3(1+1/x))^(-2/3)*(-x^-2))/x^-2 from the derivatives

You've got mismatched grouping symbols above, but I note that -x^2/x^2 equals -1.

So, I think that your answer is actually 1/3, which is the correct limit.


Here are the derivatives that I got.

numerator: \(\displaystyle -\dfrac{1}{3} \cdot \dfrac{1}{(1+1/x)^{2/3}x^2}\)

denominator: \(\displaystyle -\dfrac{1}{x^2}\)

:cool:

how I can get my equations to look 'professional'?

The math-formatting system is called LaTex. You may google for tutorials, but be warned that there are many different implementations of LaTex on the Internet; not everything that you find works here.

You may right-click any LaTex expression here, to see the coding (Show Math As >> TeX commands).

In these forums, you must enclose LaTex coding within [ֺtex] and [/ֺtex] tags.

Ciao

 

[Quaid]

wouldn't you be dividing by 0?

Not in a limit process.

1/x never equals zero, as x approaches infinity. (But, you may make 1/x as close to zero as you like, by taking x sufficiently large).

Cheers