<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] use the binomial theorem to expand   $$(x+1)^{1/3} = x^{1/3} + \frac 13 x^{-2/3}+\cdots $$  so $$ (x+1)^{1/3} - x^{1/3} = \frac 13 x^{-2/3}+\cdots \rightarrow 0\text{ as } x \to \infty.$$
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