Epsilon-delta limit definition I got $\lim_{x \rightarrow 3} \frac{x^2-2x+1}{x-2} = 4$
I need to prove that by delta epsilon. I came to delta ={1/2epsilon,1/2}
Is that right? If not can you explain me how?

Let $\forall \epsilon >0$ and we have to show that ${ \delta  }>0$ such that 
 if $0<\left| x-3 \right| <{ \delta  }$ then $\left| \frac { { x }^{ 2 }-2x+1 }{ x-2 } -4 \right| <\epsilon $ 

$$\left| \frac { { x }^{ 2 }-2x+1 }{ x-2 } -4 \right| =\left| \frac { { x }^{ 2 }-2x+1-4x+8 }{ x-2 }  \right| =\left| \frac { { x }^{ 2 }-6x+9 }{ x-2 }  \right| =\left| \frac { { \left( x-3 \right)  }^{ 2 } }{ x-3+1 }  \right|<\\\\ <\left| \frac { { \left( x-3 \right)  }^{ 2 } }{ x-3 }  \right| =\left| x-3 \right| <\delta =\epsilon  $$

