Need to solve this very simple limit $$ \lim _{x\to \infty
\:}\left(\sqrt[3]{3x^2+4x+1}-\sqrt[3]{3x^2+9x+2}\right) $$
I know how to solve these limits: by using
$a−b= \frac{a^3−b^3}{a^2+ab+b^2}$. The problem is that the standard way (not by using L'Hospital's rule) to solve this limit - very tedious, boring and tiring. I hope there is some artful and elegant solution. Thank you!
Answer
You have $$f(x)=\sqrt[3]{3x^2+4x+1}-\sqrt[3]{3x^2+9x+2}=\sqrt[3]{3x^2}\left(\sqrt[3]{1+\frac{4}{3x}+\frac{1}{3x^2}}-\sqrt[3]{1+\frac{3}{x}+\frac{2}{3x^2}}\right)$$ Using Taylor expansion at order one of the cubic roots $\sqrt[3]{1+y}=1+\frac{y}{3}+o(y)$ at the neighborhood of $0$, you get: $$f(x)=\sqrt[3]{3x^2}\left(\frac{4}{9x}-\frac{1}{x}+o\left(\frac{1}{x}\right)\right)$$ hence $$\lim\limits_{x \to \infty} f(x)=0$$
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