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

Sep 7, 2014

Let us review the definition of a limit first.

Definition ${\lim}_{x \to a} f \left(x\right) = L$ if
$\forall \epsilon > 0$, $\exists \delta > 0$ s.t.
$0 < | x - a | < \delta R i g h t a r r o w | f \left(x\right) - L | < \epsilon$

Let us now prove that ${\lim}_{x \to 5} \left(x - 1\right) = 4$.
(Note: $f \left(x\right) = x - 1$, $a = 5$, $L = 4$)
Proof
$\forall \epsilon > 0$, $\exists \delta = \epsilon > 0$ s.t.
$0 < | x - 5 | < \delta R i g h t a r r o w | \left(x - 1\right) - 4 | = | x - 5 | < \delta = \epsilon$.

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