Question

Using the ε −δ definition, show that lim 3( x -5 )=1

Answer 1

$lim_{x\rightarrow x_0}\ f(x)=A\\ \forall \epsilon>0\ \exist \delta=\delta(\epsilon):\ \forall x\ 0<|x-x_0|<\delta\\ |f(x)-A|<\epsilon\text{ (Cauchy definition)}\\ lim_{x\rightarrow 16/3}\ 3(x-5)=1\\ \forall \epsilon>0\ \exist \delta=\epsilon/3:\ \forall x\ 0<|x-16/3|<\delta\\ |3(x-5)-1|<\epsilon\\ |3(x-5)-1|=|3x-16|=3|x-16/3|\\ |x-16/3|=\frac{1}{3}|3(x-5)-1|\\ \delta=\epsilon/3$.