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sluggerbroth's Question from Math Help Forum

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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

Administrator
Staff member
Jan 26, 2012
4,052
I wonder if this question requires a delta-epsilon proof instead? I always disliked those!
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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.