Find limit of $\sqrt[3]{n+1}-\sqrt[3]{n}$

Question

I have to find the limit of: $$\sqrt[3]{n+1}-\sqrt[3]{n}$$

What I know is that multiplying can be the key here, so I simplified it to $$\frac{1}{(n+1)^{\tfrac{2}{3}}+n^{\tfrac{1}{3}}(n+1)^{\tfrac{1}{3}}+n^{\tfrac{2}{3}}}.$$

Now what I do not understand: I know that this converges to zero, since all the terms of the denominator grow as $n$ increases. But how do I formally finish this proof?

Wouldn't I have to show for each term that it grows continually?

I am not quite sure how to do this. Any hints would be very kindly appreciated, thanks!

Answer 1

Hint. We have that $$0<\sqrt[3]{n+1}-\sqrt[3]{n}=\frac{1}{(n+1)^{\tfrac{2}{3}}+n^{\tfrac{1}{3}}(n+1)^{\tfrac{1}{3}}+n^{\tfrac{2}{3}}}\leq \frac{1}{n^{2/3}}$$ and then use the Squeeze Theorem.

Answer 2

An alternative approach to Robert Z's squeeze theorem answer:

$$n\le n+1=n\left(1+{1\over n}\right)\le n\left(1+{3\over n}+{3\over n^2}+{1\over n^3}\right)=n\left(1+{1\over n}\right)^3$$

so

$$\sqrt[3]n\le\sqrt[3]{n+1}\le\sqrt[3]n\left(1+{1\over n}\right)$$

so

$$0\le\sqrt[3]{n+1}-\sqrt[3]n\le{\sqrt[3]n\over n}={1\over n^{2/3}}$$

Answer 3

we have

$$\sqrt[3]{n+1}-\sqrt[3]{n}=$$

$$n^\frac 13 \left((1+\frac 1 n)^\frac 13 -1 \right)=$$

$$n^\frac 13\left(1+\frac{1}{3n}(1+\epsilon(n))-1\right)=$$

$$\frac{1}{3n^\frac 23}(1+\epsilon(n))$$

the limit is $0$.