<s>[INST] Finding the limit $\lim_{x\rightarrow \infty} \sqrt[3]{x+1}-\sqrt[3]{x}$ I am trying to  find this limit
$$\lim_{x\rightarrow \infty} \sqrt[3]{x+1}-\sqrt[3]{x}$$
My so far method is this


*

*$f(x)>0.$

*$f^{\prime}(x)=\frac{1}{3\sqrt[3]{(x+1)^2}}-\frac{1}{3\sqrt[3]{x^2}}<0.$

*For every $0<y<1$ the equation $f(x)=y$ has a unique solution(found in maple):


$$x=\frac{1}{3}\cdot \frac{\frac{1}{3}y^2(3y^2+\sqrt{-3y^4+12y})+\frac{1}{6}\frac{3y^2+\sqrt{-3y^4+12y}}{y}-y^4-2y}{y}$$
The previous facts implies that the limit is $0$.
I am wondering if there is any way easier than this to find the limite.
thanks in advance.
 [/INST] Yet another method:  let $a = (x+1)^{1/3}$, $b = x^{1/3}$, so $$f(x) = (x+1)^{1/3} - x^{1/3} = a - b = \frac{(a-b)(a^2 + ab + b^2)}{a^2 + ab + b^2} = \frac{a^3 - b^3}{a^2 + ab + b^2}.$$  But $a^3 - b^3 = (x+1) - x = 1$, hence $$f(x) = \frac{1}{a^2 + ab + b^2}.$$  Now since $a \to \infty$ and $b \to \infty$ as $x \to \infty$, it is obvious that $$\lim_{x \to \infty} f(x) = 0.$$
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