Find the following limit $$\lim_{x \to \infty} \,\, \sqrt[3]{x^3-1} – x – 2$$
How do I find this limit? If I had to guess I'd say it converges to $-2$ but the usual things like L'Hôpital or clever factorisation don't seem to work in this case.
[Math] Evaluate: $\lim_{x \to \infty} \,\, \sqrt[3]{x^3-1} – x – 2$
calculuslimits
Best Answer
We have using Taylor series
$$(x^3-1)^{\frac13}-x-2=x\left(1-\frac1{x^3}\right)^{\frac13}-x-2\sim_\infty x\left(1-\frac1{3x^3}\right)-x-2\\=-\frac1{3x^2}-2\xrightarrow{x\to\infty}-2$$