Originally Posted by

**Showcase_22** I would just like to point out that i've never done a question like this before so here's my attempt:

$\displaystyle f_n(x)=\frac{2nx}{n^p}+x^2=2n^{1-p}x+x^2$

Suppose $\displaystyle p>2$.

Need to find $\displaystyle \delta>0$ such that $\displaystyle |x-y|<\delta \Rightarrow |f(x)-f(y)|<\epsilon$.

Let $\displaystyle x=\frac{\delta}{2}$ and $\displaystyle y=\frac{\delta}{3}$ and fix $\displaystyle \delta=\min \{ 1, \epsilon \}$

$\displaystyle |2n^{1-p}x+x^2-2n^{1-p}y-y^2|=|2n^{1-p}(x-y)+(x^2-y^2)| \leq |2n^{1-p}(x-y)|+|x^2-y^2|$

But:

$\displaystyle =|2n^{1-p}(x-y)|+|(x+y)(x-y)|< |2n^{1-p}|\delta+|x+y|\delta=|2n^{1-p}|\delta+\left| \frac{5\delta}{6} \right|\delta$

But since $\displaystyle 0<\delta<1 \Rightarrow \frac{5 \delta^2}{6} \leq \frac{5 \delta}{6}$.

Hence we have $\displaystyle |f(x)-f(y)| < |2n^{1-p}| \delta+\frac{5 \delta}{6}=\delta \left( |2n^{1-p}|+\frac{5}{6} \right)$

Hence define $\displaystyle \epsilon=\delta \left( |2n^{1-p}|+\frac{5}{6} \right)$

EDIT: I would really like to point out that i'm not sure if this is right. I haven't really encountered uniform continuity before!! (Worried)

(I also can't see where i'm used the property that p>2....)