Problems solving a limit which results in an indeterminate form

[greg_rack]

Homework Statement

$\lim_{x \to +\infty}(\sqrt[3]{x^3-4x^2}-x)$

Relevant Equations

none

Hi guys, I am having difficulties in solving this limit.

Below, I'll attach my procedure which ends up in the indeterminate form $0\cdot \infty$...
How could I solve it?

$$\lim_{x \to +\infty}(\sqrt[3]{x^3-4x^2}-x) \rightarrow
\lim_{x \to +\infty}(x\sqrt[3]{1-\frac{4}{x}}-x) \rightarrow
\lim_{x \to +\infty}[x(\sqrt[3]{1-\frac{4}{x}}-1)]$$

 

[etotheipi]

What is the binomial expansion of $(1-\frac{4}{x})^{1/3}$?

 

[PeroK]

Homework Statement:: $\lim_{x \to +\infty}(\sqrt[3]{x^3-4x^2}-x)$
Relevant Equations:: none

Hi guys, I am having difficulties in solving this limit.

Below, I'll attach my procedure which ends up in the indeterminate form $0\cdot \infty$...
How could I solve it?

$$\lim_{x \to +\infty}(\sqrt[3]{x^3-4x^2}-x) \rightarrow
\lim_{x \to +\infty}(x\sqrt[3]{1-\frac{4}{x}}-x) \rightarrow
\lim_{x \to +\infty}[x(\sqrt[3]{1-\frac{4}{x}}-1)]$$

You can do it using the binomial expansion, as above. Or, you use a general factorisation method using:
$$(f(x)^{1/3} - x)(f(x)^{2/3} + xf(x)^{1/3} + x^2) = f(x) - x^3$$ with $f(x) = x^3 -4x^2$

 

[Mark44]

You can do it using the binomial expansion, as above. Or, you use a general factorisation method using:
$$(f(x)^{1/3} - x)(f(x)^{2/3} + xf(x)^{1/3} + x^2) = f(x) - x^3$$ with $f(x) = x^3 -4x^2$

This is the tack I would take. The basic ideas when dealing with a difference of square roots or a difference of cube roots (as in this problem) are these identities:
$(a - b)(a + b) = a^2 - b^2$
$(a - b)(a^2 + ab + b^2) = a^3 - b^3$
$(a +b)(a^2 - ab + b^2) = a^3 + b^3$
In the first identity, a or b stands for the square root of some expression.
In the second and third, a or b stands for the cube root of some expression.

 

[greg_rack]

Great, thanks!