Question

I've found out that the limit of the equation below is -5.

But I'm not sure how to start when proving that limit with the epsilon-delta definition?

Answer 1

\[\lim_{x \rightarrow 2}(-5)\]

Answer 2

the limit of a constant is the constant ...

if you really want, let the equation be -5 + 0x

Answer 3

find the largest d such that: 0<|x-2|<d given 0< |f(x) - L| < e

Answer 4

-e < f(x) - L < e

L-e < f(x) < L+e

convert the range into the domain

-5-e < -5+0x < -5+e

-e < 0x < e

i would say this is true for all values of x

but im rusty on my e,d definitions

Answer 5

what is your f(x)?

Answer 6

f(x)= -5 ?

Answer 7

yeah, thats what it looks like from the posting, im jsut not sure if this is starting i the middle of a question where you found the limit as -5 or if its the start of it. it feels like a middle of the line setup

Answer 8

I'm always used to working with regular equations like 7x+70 for example. But this problem seemed unfamiliar to me.

Answer 9

graphically, its a sinch



delta = 0 to infinity, an we remain within some epsilon of f(x)

Answer 10

right

Answer 11

I want to try to get to absolute value of f(x) - L < e

Answer 12

so show me how you would do:

f(x) = 2x + 1, as x approaches 2

Answer 13

as "x" approaches 2, the limit would be 5
\[\left| 2x+1 \right|-5<\epsilon \]

Answer 14

|f(x)-5| not |f(x)| - 5, or am i mistaken?

Answer 15

\[\left| 2x-4 \right|<\epsilon\]

Answer 16

\[\left| 2(x-2) \right|<\epsilon\]

Answer 17

\[\left| 2 \right|*\left| (x-2) \right|<\epsilon/2\]

Answer 18

|x-2| < e/2 but yeah

Answer 19

right

Answer 20

-e/2 < x-2 < e/2

(4-e)/2 < x < (4+e)/2

is valid for all x, and if we know an epsilon, we can easily pick delta for it. right?

Answer 21

how does the epsilon-delta help us define the limit ... as x to 2?

Answer 22

does our definition say \[|f(x)-L|<\epsilon\]
or can epsilon not equate to zero? its been awhile

Answer 23

yep. that's it. that's the form I want to get it into to finish verifying the limit.

Answer 24

http://www.quora.com/What-is-an-epsilon-delta-proof-of-the-Constant-Rule-of-Limits

this is as proofy as i can find for it

Answer 25

f(x) + 5 is 0 which is less than all epsilon greater than 0

Answer 26

thanks for the link. I will look into that

Answer 27

other than that, im just not in practice to be able to express it that well.

Answer 28

good luck tho

Answer 29

O ok. thanks for at least trying, bud.

I'll figure it out