- #1
misogynisticfeminist
- 370
- 0
This is a worked example in the book, but i can't figure out why it is this way.
Using [tex] 0 < (x-a) < \delta [/tex] and [tex] (f(x)-L) < \epsilon [/tex] for [tex] lim_(x\rightarrow a) f(x) = L [/tex].
Prove that [tex] lim_{x\rightarrow2} (4x-3) = 5 [/tex]
I can understand the steps until
[tex] {4} (x-2) < \epsilon [/tex] and [tex] 0 < (x-2) < \delta [/tex]
then, they suddenly get, [tex] \epsilon = \delta/4 [/tex]...
Cant really understand how they arrive at there, and they don't even show that the the limit is 5.
I'm self taught too, so my understanding might be a little wrong.
Using [tex] 0 < (x-a) < \delta [/tex] and [tex] (f(x)-L) < \epsilon [/tex] for [tex] lim_(x\rightarrow a) f(x) = L [/tex].
Prove that [tex] lim_{x\rightarrow2} (4x-3) = 5 [/tex]
I can understand the steps until
[tex] {4} (x-2) < \epsilon [/tex] and [tex] 0 < (x-2) < \delta [/tex]
then, they suddenly get, [tex] \epsilon = \delta/4 [/tex]...
Cant really understand how they arrive at there, and they don't even show that the the limit is 5.
I'm self taught too, so my understanding might be a little wrong.
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