Question

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

Answer 1

Given any $\epsilon>0$ we choose $\delta_{\epsilon}<\dfrac{\epsilon}{3}$

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

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

we evaluate the difference:

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

and we can see that-

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

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

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

Which prove the above point.