How to find the limit of a function with a third root?

Question

I have the function $x^2(\sqrt[3]{x^3+1} - x) $ and have to find the limit for $x \rightarrow \infty $.

After many hours of forming around I still have no clue how to find it. Is there anybody who could give me a hint?

Answer 1

Hint: Use the identity $a^3-b^3=(a-b)(a^2+ab+b^2)$ with $a=\sqrt[3]{x^3+1}$ and $b=x$ to rationalize the expression.

$$\begin{align} x^2(\sqrt[3]{x^3+1}-x) & =x^2(\sqrt[3]{x^3+1}-x)\frac{\sqrt[3]{(x^3+1)^2}+x\sqrt[3]{x^3+1}+x^2}{\sqrt[3]{(x^3+1)^2}+x\sqrt[3]{x^3+1}+x^2}\\ & ={x^2\over \sqrt[3]{(x^3+1)^2}+x\sqrt[3]{x^3+1}+x^2} \end{align}$$

Note that $a^3-b^3=1$ when $a=\sqrt[3]{x^3+1}$ and $b=x$.

Now, multiply top and bottom by $1/x^2$ and take the limit.

Answer 2

Another solution would consist in writing $$\sqrt[3]{x^3+1} - x=x \sqrt[3]{1+\frac 1{x^3}}-x=x (\sqrt[3]{1+\frac 1{x^3}}-1) $$ which makes the initial expression $$A=x^2(\sqrt[3]{x^3+1} - x)=x^3 (\sqrt[3]{1+\frac 1{x^3}}-1)$$ Now, use Taylor series for $\sqrt[3]{1+{y}}$ built at $y=0$ $$\sqrt[3]{1+{y}}=1+\frac{y}{3}-\frac{y^2}{9}+O\left(y^3\right)$$ and replace $y=\frac 1{x^3}$. So, back to the expression $$A\approx x^3(1+\frac{1}{3x^3}-\frac{1}{9x^6}-1)=\frac{1}{3}-\frac{1}{9x^3}$$ which gives both the limit and how it is approached.

If you plot on the same graph the initial function and the approximation, you should be amazed to notice how much the two curves are close to eachother. Computing for $x=10$ (quite far away from $\infty$), the value of the initial function is $\approx 0.3332222839$ while the approximation gives $\approx 0.3332222222$.