Limit of : $\lim\limits_{n\to\infty}{(\sqrt[3]{n+1}-\sqrt[3]{n})}$.

Question

I have to find $$\lim_{n\to\infty}{(\sqrt[3]{n+1}-\sqrt[3]{n})}$$ here what I did $$\lim_{n\to\infty}{(\sqrt[3]{n+1}-\sqrt[3]{n})}$$=$$\lim_{n\to\infty}{\left(\sqrt[3]{n\left(1+\frac{1}{n}\right)}-\sqrt[3]{n}\right)}$$ How can I proceed?

Answer 1

$$\lim_{n\to\infty}{(\sqrt[3]{n+1}-\sqrt[3]{n})}$$$$=\lim_{n\to\infty}\frac{(n+1)-n}{(n+1)^{\frac{2}{3}}+(n+1)^{\frac{1}{3}}n^{\frac{1}{3}}+n^{\frac{2}{3}}}$$$$=\lim_{n\to\infty}\frac{1}{(n+1)^{\frac{2}{3}}+(n+1)^{\frac{1}{3}}n^{\frac{1}{3}}+n^{\frac{2}{3}}}$$$$=0.$$

Answer 2

You may Taylor-expand

$$\sqrt[3]{n \left(1+\frac{1}{n}\right)}-\sqrt[3]{n}=\sqrt[3]{n}\left(1+\frac{1}{n}\right)^{1/3}- \sqrt[3]{n}=\sqrt[3]{n}\left(1+\frac{1}{3n}\right)- \sqrt[3]{n} =\frac{\sqrt[3]{n}}{3n}$$

and then take the limit to get zero.

Answer 3

My suggestion is to consider $$ f(x)=\frac{\sqrt[3]{1+x^3}-1}{x} $$ so your limit is $$ \lim_{n\to\infty}f(1/\sqrt[3]{n}) $$ and you solve the problem if you find that $$ \lim_{x\to0}f(x) $$ exists, because your limit would be the same. This limit is much easier: $$ \lim_{x\to0}\frac{\sqrt[3]{1+x^3}-1}{x} $$ is the derivative at $0$ of $g(x)=\sqrt[3]{1+x^3}$; since $$ g'(x)=\dfrac{3x^2}{3\sqrt[3]{(1+x^3)^2}} $$ we have $g'(0)=0$. So you conclude that $$ \lim_{n\to\infty}\bigl({\textstyle\sqrt[3]{n+1}-\sqrt[3]{n}}\bigr)=0 $$

How did I find $f$? Just formally substituting $n=1/x^3$.

Answer 4

By binomial approximation

$$\sqrt[3]{n+1}=\sqrt[3]n\left(1+\frac1n\right)^\frac13\approx \sqrt[3]n\left(1+\frac1{3n}\right)=\sqrt[3]n+\frac{\sqrt[3]n}{3n}$$

therefore

$$\sqrt[3]{n+1}-\sqrt[3]{n}\approx \sqrt[3]n+\frac{\sqrt[3]n}{3n}-\sqrt[3]n=\frac{\sqrt[3]n}{3n} \to 0$$

Answer 5

$(1+1/n)^{1/3} \le 1+(1/3)(1/n)$ (Bernouilli inequality)

$0<n^{1/3}((1+1/n)^{1/3}-1) \le$

$n^{1/3}(1/3)(1/n)\le n^{-2/3}$.

Squeeze.

https://en.m.wikipedia.org/wiki/Bernoulli%27s_inequality

Answer 6

Notice that

$$\left(m+\frac1{3m^2}\right)^3=m^3+1+\frac1{3m^3}+\frac1{27m^6}>m^3+1$$

so that

$$m<\sqrt[3]{m^3+1}<m+\frac1{3m^2}$$ and by letting $n=m^3$, the requested limit squeezes to $0$.