Given any ϵ>0 we choose δϵ<3ϵ
we have to prove that given any ϵ>0 we can find a δϵ such that
∣f(x)−1∣<ϵ for x∈(2−δϵ,2+δϵ)
we evaluate the difference:
∣f(x)−1∣=∣3x−5−1∣=∣3x−6∣=3∣x−2∣
and we can see that-
∣f(x)−1∣<ϵ⇒∣x−2∣<3ϵ
So, given ϵ>0 we can choose δϵ and we have:
x∈(2−δϵ,2+δϵ)⇒∣x−2∣<3ϵ⇒∣f(x)−1∣<ϵ
Which prove the above point.
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