> 
> $$(\forall\varepsilon>0)(\exists K)(\forall x)(x>K\Rightarrow\frac{1}{4x}<\varepsilon)$$
> 
> 
> 

Here is my attempted proof: Let $\varepsilon>0$ be a real number. Let $K$ be $\frac{1}{2\varepsilon}$. Assume $x$ is a real number and $x>K$. Then $x>\frac{1}{2\varepsilon}$, so $x>\frac{1}{4\varepsilon}$. Therefore, $4x\varepsilon>1$, so $\frac{1}{4x}<\varepsilon$.

Is it correct? According to my pen friend who is a math major there's a fault in it but he didn't tell me which, leaving me to figure it out, but I couldn't

**EDIT:** $K$ is a natural number, I think that's the error. So do you have any ideas for a better proof?