Calculate this limit : $\lim_{x \rightarrow -\infty} -x-\sqrt[3]{-x^3-x}$

Question

Calculate this limit : $$\lim_{x \rightarrow -\infty} -x-\sqrt[3]{-x^3-x}$$

I tried to factorize but i can't find any result

Answer 1

Hint :Use that $(1+x)^{\frac{1}{3}} = 1 + \frac{x}{3} - \frac{x^2}{9} + O(x^3)$.

EDIT As user doesn't seem to know Taylor series and Big O notation, the following approach might be considered: For sake of simplicity, say $X = -x$, such that we're interested in:

$\lim_{X \rightarrow +\infty} X - (X^3 + X)^{\frac{1}{3}}$

Write $X - (X^3 + X)^{\frac{1}{3}} = X(1 - (1 + \frac{1}{X^2})^{\frac{1}{3}})$, and notice that we have the following inequalities because $1 + \frac{1}{X^2} > 1$, that:

$X(1 - (1 + \frac{1}{X^2})^{\frac{1}{2}}) \leq X(1 - (1 + \frac{1}{X^2})^{\frac{1}{3}}) \leq X(1 - (1 + \frac{1}{X^2})^{\frac{1}{4}})$

Now using that $1 - x^{\frac{1}{2}} = \frac{(1 - x^{\frac{1}{2}})(1 + x^{\frac{1}{2}})}{1 + x^{\frac{1}{2}}} = \frac{1 - x}{1 + x^{\frac{1}{2}}}$, we have the following equivalence

$\iff X\frac{-\frac{1}{X^2}}{1 + (1 + \frac{1}{X^2})^{\frac{1}{2}}} \leq X(1 - (1 + \frac{1}{X^2})^{\frac{1}{3}}) \leq X\frac{1 - (1 + \frac{1}{X^2})^{\frac{1}{2}}}{1 + (1 + \frac{1}{X^2})^{\frac{1}{4}}}$

$\iff X\frac{-\frac{1}{X^2}}{1 + (1 + \frac{1}{X^2})^{\frac{1}{2}}} \leq X(1 - (1 + \frac{1}{X^2})^{\frac{1}{3}}) \leq X\frac{-\frac{1}{X^2}}{(1 + (1 + \frac{1}{X^2})^{\frac{1}{4}})(1 + (1 + \frac{1}{X^2})^{\frac{1}{2}})}$

$\iff \lim_{X \rightarrow +\infty}\frac{-\frac{1}{X}}{1 + (1 + \frac{1}{X^2})^{\frac{1}{2}}} \leq \lim_{X \rightarrow +\infty}X(1 - (1 + \frac{1}{X^2})^{\frac{1}{3}}) \leq \lim_{X \rightarrow +\infty}\frac{-\frac{1}{X}}{(1 + (1 + \frac{1}{X^2})^{\frac{1}{4}})(1 + (1 + \frac{1}{X^2})^{\frac{1}{2}})}$

and therefore yielding that $\lim_{X \rightarrow +\infty}X(1 - (1 + \frac{1}{X^2})^{\frac{1}{3}}) = 0$ thanks to squeeze theorem.

Answer 2

Hint: $$-x-\sqrt[3]{-x^3-x}=(\sqrt[3]{x^3+x}-x)\frac{(\sqrt[3]{x^3+x})^2+x\sqrt[3]{x^3+x}+x^2}{(\sqrt[3]{x^3+x})^2+x\sqrt[3]{x^3+x}+x^2}=\frac{x}{(\sqrt[3]{x^3+x})^2+x\sqrt[3]{x^3+x}+x^2}$$