Question

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

Answer 1

We have to prove that given any $ε>0$  we can find a $δ_ε>0$  such that

$| f ( x ) − 1 | < ε \quad \forall x ∈ ( 2 − δ_ ε , 2 + δ_ ε )$

Let's evaluate the difference

$| f ( x ) − 1 | = | 3 x − 5 − 1 | = | 3 x − 6 | = 3 | x − 2 |$

Therefore

$| f ( x ) − 1 | < ε ⇔ | x − 2 | < \frac{ε}{3}$

Given, that $ε>0$ we can choose $δ_ε<\frac{ε}{3}$

Hence

$x ∈ ( 2 − δ_ ε , 2 + δ_ ε ) ⇒ | x − 2 | < \frac{ε}{3} ⇔ | f ( x ) − 1 | < ε$