Question

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?

Answer 1

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!

Answer 2

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

Answer 3

lim{x->6} (2x+1) = 13

f(d) = 2d+1 = 13+e

2d = 12+e

d = (12+e)/2 i think

Answer 4

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

Answer 5

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

Answer 6

d < e/2 is good

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

|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