## anonymous 5 years ago Calculate the value of delta that corresponds to epsilon given limx->6 (2x+1) = 13, according to the definition of limits. Why delta = epsilon/2 is the only suitable solution? I have two cases: 1. If delta > epsilon/2 that's not going to work because our original |f(x)-L| will be greater than epsilon which is not possible: |f(x)-L| > epsilon 2. If delta < epsilon / 2 then |f(x)-L| < epsilon - a (where a < epsilon) which is ok since |f(x)-L| < epsilon - a < epsilon. So the first case is wrong. But what's about the second case? What if delta < epsilon/2? Will it still be wrong? Why?

1. Owlfred

Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

2. amistre64

since this is a line; your delta is simple. :)

3. amistre64

lim{x->6} (2x+1) = 13 f(d) = 2d+1 = 13+e 2d = 12+e d = (12+e)/2 i think

4. amistre64

maybe.... es and ds tend to confuse me lol

5. amistre64

or should that be f(x+d)??

6. amistre64

d < e/2 is good

7. amistre64

i can see it, but trying to describe it eludes me....

8. anonymous

From formal definition of limit: $\lim_{x \rightarrow a}f(x) = L$ $|f(x) - L| < \epsilon$ So we have: $|2x+1-13| < \epsilon$ $|x-6| < \epsilon/2$ And also from formal definition of limit: $0 < |x - a| < \delta$ Which gives us: $0 < |x - 6| < \delta$ So delta can be anywhere between 0 and epsilon/2

9. anonymous

I would say that u can go to Paul's Online Math Notes for additional help it's a pretty good math website yall

10. myininaya

|2x+1-13|=|2x-12|=2*|x-6|<e solvinf this for |x-6| we can what we need to make d so |x-6|<e/2 so d=e/2