# sluggerbroth's Question from Math Help Forum

#### Sudharaka

##### Well-known member
MHB Math Helper
Title: Prove this limit

$\lim_{x\rightarrow 1}(x^2-2x+4)=3$
Hi sluggerbroth,

\begin{eqnarray}

\lim_{x\rightarrow 1}(x^2-2x+4)&=&\lim_{x\rightarrow 1}x^2-2\lim_{x\rightarrow 1}x+\lim_{x\rightarrow 1}4\\

&=&1^2-2+4\\

&=&3

\end{eqnarray}

Kind Regards,
Sudharaka.

#### Jameson

Staff member
I wonder if this question requires a delta-epsilon proof instead? I always disliked those!

#### Sudharaka

##### Well-known member
MHB Math Helper
I wonder if this question requires a delta-epsilon proof instead? I always disliked those!
Exactly. This is something that didn't occur to me.

Let $$f(x)=x^2-2x+4\mbox{ and }l=3$$. Take any $$\epsilon>0$$, and consider, $$|f(x)-l|$$

\begin{eqnarray}

|f(x)-l|&=&|x^2-2x+4-3|\\

&=&|x^2-2x+1|\\

&=&(x-1)^2\\

&<&\epsilon\mbox{ whenever }|x-1|<\sqrt{\epsilon}\\

\end{eqnarray}

Take $$\delta=\sqrt{\epsilon}$$ and we get,

$|f(x)-l|<\epsilon\mbox{ whenever }|x-1|<\delta$

Therefore for every $$\epsilon>0$$ there exists a $$\delta>0$$ such that,

$|f(x)-l|<\epsilon\mbox{ whenever }|x-1|<\delta$

Hence,

$\lim_{x\rightarrow 1}f(x)=l$

$\Rightarrow\lim_{x\rightarrow 1}(x^2-2x+4)=3$

Kind Regards,
Sudharaka.