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?

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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?

Mathematics
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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

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maybe.... es and ds tend to confuse me lol
or should that be f(x+d)??
d < e/2 is good
i can see it, but trying to describe it eludes me....
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
I would say that u can go to Paul's Online Math Notes for additional help it's a pretty good math website yall
|2x+1-13|=|2x-12|=2*|x-6|

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