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?

Hey! We 've verified this expert answer for you, click below to unlock the details :)

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

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

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.