Problem in calculating limit with cube root

Question

I need to calculate the next limit: $\lim_{x \to \infty}(\sqrt[3]{x^3+2}-\sqrt[3]{x^3+1})$

Answer 1

Hint

$$a-b = \frac{\left(a-b\right)\left(a^2+ab+b^2\right)}{a^2+ab+b^2}= \frac{a^3-b^3}{a^2+ab+b^2}$$

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You can use this with $a=\sqrt[3]{x^3+2}$ and $b=\sqrt[3]{x^3+1}$ to get rid of the cube roots (in the numerator).

Answer 2

HINT

As an alternative by binomial first order approximation

• $\sqrt[3]{x^3+2}=x(1+2/x^3)^\frac13\sim x\left(1+\frac2{3x^3}\right)=x+\frac2{3x^2}$
• $\sqrt[3]{x^3+1}=x(1+1/x^3)^\frac13\sim x\left(1+\frac1{3x^3}\right)=x+\frac1{3x^2}$