Prove (only with definition) that $f(x)=\frac{x^3+1}{x^2}, x>0$ is not uniformly continuous

Question

I want to prove (only with definition) that $f(x)=\dfrac{x^3+1}{x^2}, x>0$ is not uniformly continuous. Obviously, $f$ has the problem near at zero. I suppose that $f$ is uniformly continuous. So, for all $\epsilon >0$, exists some $\delta>0$, such that if $x,y>0, |x-y|<\delta\Rightarrow |f(x)-f(y)|<\epsilon$. I tried to take some $0<x<\delta$ and $y=x/2$. Therefore, $|x-y|<\delta$, and I want to prove that $|f(x)-f(y)| $ is bigger than a fix number $\epsilon >0$. However,it seems that this choice about $x$ and $ y$ did not work. Any ideas, what's a better choice for $x$ and $ y$? Thanks

Answer 1

Take $\varepsilon=1$ and let $\delta$ be an arbitrary number greater than $0$. Take any number $x\in(0,\delta)$. Since $\lim_{x\to0^+}f(x)=\infty$, there is a $y\in(0,\delta)$ such that $f(y)\geqslant f(x)+1$. So, $|x-y|<\delta$ and $\bigl|f(x)-f(y)\bigr|\geqslant1=\varepsilon$.