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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?
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?

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Owlfred
 5 years ago
Best ResponseYou've already chosen the best response.0Hoot! 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!

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0since this is a line; your delta is simple. :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0lim{x>6} (2x+1) = 13 f(d) = 2d+1 = 13+e 2d = 12+e d = (12+e)/2 i think

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0maybe.... es and ds tend to confuse me lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0or should that be f(x+d)??

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i can see it, but trying to describe it eludes me....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0From formal definition of limit: \[\lim_{x \rightarrow a}f(x) = L\] \[f(x)  L < \epsilon\] So we have: \[2x+113 < \epsilon\] \[x6 < \epsilon/2\] 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I would say that u can go to Paul's Online Math Notes for additional help it's a pretty good math website yall

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.02x+113=2x12=2*x6<e solvinf this for x6 we can what we need to make d so x6<e/2 so d=e/2
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