Argue that $f$ is not uniformly continuous

Question

Let $f(x)=x^3$. We want to show that $f$ is not uniformly continuous.

Can someone please explain from $|f(x_n)-f(y_n)=\big|(n+\frac{1}{n})^3 - n^3)\big| = 3n+\displaystyle\frac{3}{n}$? How does $\big|(n+\frac{1}{n})^3-n^3\big| $ yield $3n+\displaystyle\frac{3}{n}$? ${}{}{}{}$

Answer 1

The first part is just the Binomial Theorem: $$ \left(n + \frac{1}{n}\right)^3 = n^3 + 3 n^2 \cdot \frac{1}{n} + 3 n \frac{1}{n^2} + \frac{1}{n^3} = n^3 + 3n + \frac{3}{n} + \frac{1}{n^3} $$ So $$ \left(n + \frac{1}{n}\right)^3 - n^3 = 3n + \frac{3}{n} + \frac{1}{n^3} $$ The author seems to skip the $\frac{1}{n^3}$ term, but no matter. The expression is still $>3$, which is the point.

Answer 2

The algebraic identity $$a^3-b^3=(a-b)(a^2+ab+b^2)$$

Answer 3

You're missing the $\frac{1}{n^3}$ term. Indeed, $$\left(n+\frac{1}{n}\right)^3-n^3=n^3+3n^2\frac{1}{n}+3n\frac{1}{n^2}+\frac{1}{n^3}-n^3=3n+3\frac{1}{n}+\frac{1}{n^3}$$