How do you use the epsilon delta definition of limit to prove that $lim_(x->5)(x-1)= 4$ ?

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How do you use the epsilon delta definition of limit to prove that $lim_(x->5)(x-1)= 4$ ?

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Answer 1 · 2014-09-07T13:27:20

Let us review the definition of a limit first.

Definition $lim_{x to a}f(x)=L$ if
$forall epsilon > 0$ , $exists delta > 0$ s.t.
$0<|x-a|< delta Rightarrow |f(x)-L| < epsilon$

Let us now prove that $lim_{x to 5}(x-1)=4$ .
(Note: $f(x)=x-1$ , $a=5$ , $L=4$ )
Proof
$forall epsilon>0$ , $exists delta=epsilon>0$ s.t.
$0<|x-5|< delta Rightarrow |(x-1)-4|=|x-5|< delta= epsilon$ .

Remark: The key is to find $delta$ in terms of $epsilon$ . In this particular proof above, we can set $delta=epsilon$ .

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How do you use the epsilon delta definition of limit to prove that $lim_(x->5)(x-1)= 4$ ?

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How do you use the epsilon delta definition of limit to prove that lim_(x->5)(x-1)= 4lim_(x->5)(x-1)= 4 ?

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How do you use the epsilon delta definition of limit to prove that $lim_(x->5)(x-1)= 4$ ?