Question

need help on epsilon delta theorem

Answer 1

i got up to epsilon = 4delta but i'm not sure what to do next

Answer 2

(e is epsilon and d is delta)
so we have
\[\lim_{x \rightarrow 1}\frac{2+4x}{3}=2\]
so first thing to do is find what to let delta be
So you said you have played with |f(x)-L|<e until you seen |x-a|<d

Answer 3

a is 1 and L is 2 by the way

Answer 4

you can multiply the 3 and do |(2+4x)-6|<e right?

Answer 5

well you would have |(2+4x)-6|<3e

Answer 6

multiply one side by 3 multiply the other side by 3

Answer 7

so if you get e=4/3*d is the proof over?

Answer 8

So well finding delta isn't the same as the proof part

Answer 9

But the hard part is over.

Answer 10

oh, how do you finish the proof

Answer 11

\[\text{ Let } \epsilon>0 \text{ be given .} \text{Choose } \delta =\frac{3}{4} \epsilon . \text{ Then whenever } 0<|x-1|<\delta \\ \text{ we have } |f(x)-L|=|\frac{2+4x}{3}-2|=.. \]

Answer 12

see if you can finish the proof

Answer 13

isn't that just equal to epsilon?

Answer 14

we want to show this is less than epsilon using our delta we chose

Answer 15

\[|f(x)-L|=|\frac{2+4x}{3}-2|=|\frac{2+4x-6}{3}|=\frac{1}{3}|2+4x-6|=?\]

Answer 16

1/3e

Answer 17

how did you get that?

Answer 18

wait is it like |(2+4x)-6|<4/3d=e

Answer 19

\[|f(x)-L|=|\frac{2+4x}{3}-2|=|\frac{2+4x-6}{3}| \\ =\frac{1}{3}|2+4x-6|=\frac{1}{3}|4x-4|=\frac{4}{3}|x-1|<\frac{4}{3} \delta\]
and yeah since we chose delta to be 3 e/4 then we have shown this is epsilon for a chosen delta

Answer 20

this is less than epsilon for a *

Answer 21

i have another question sorry, how would you find the delta if you had
\[\lim_{x \rightarrow 3}x^3=9\]