Question #180423

Using the epsilon minus delta definition,show that lim X tends to 2 (3x-5)=1


1
Expert's answer
2021-04-29T17:19:28-0400

Given any ϵ>0\epsilon>0 we choose δϵ<ϵ3\delta_{\epsilon}<\dfrac{\epsilon}{3}


we have to prove that given any ϵ>0\epsilon>0 we can find a δϵ\delta_{\epsilon} such that


f(x)1<ϵ for x(2δϵ,2+δϵ)|f(x)-1|<\epsilon \text{ for } x\in (2-\delta_{\epsilon},2+\delta_{\epsilon})


we evaluate the difference:

f(x)1=3x51=3x6=3x2|f(x)-1|=|3x-5-1|=|3x-6|=3|x-2|


and we can see that-


f(x)1<ϵx2<ϵ3|f(x)-1|<\epsilon \Rightarrow |x-2|<\dfrac{\epsilon}{3}


So, given ϵ>0\epsilon>0 we can choose δϵ\delta_{\epsilon} and we have:


x(2δϵ,2+δϵ)x2<ϵ3f(x)1<ϵx\in (2-\delta_{\epsilon},2+\delta_{\epsilon})\Rightarrow |x-2|<\dfrac{\epsilon}{3}\Rightarrow |f(x)-1|<\epsilon


Which prove the above point.


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