Epsilon-delta limit definition

Question

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?

Answer 1

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 $

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

Answer 2

|$\frac {(x-3)(x-3)}{x-2}|$< $\epsilon$

$\delta$ <= $\frac{1}{2}$

|x-3|<$\delta$ <=$\frac{1}{2}$

-$\frac{1}{2}$ < x-3 < $\frac{1}{2}$

$\frac{1}{2}$ < x-2 < $\frac{3}{2}$

2>$\frac {1}{x-2}$>$\frac{2}{3}$

$\delta$ = min{$\frac{1}{2}$$\epsilon$, $\frac{1}{2}$}

|$\frac {(x-3)(x-3)}{x-2}|$ < $\delta$*2 = $\epsilon$