Question

Consider the line f(x)=mx+b, where m does not equal 0, how do you use the epsilon delta definition of a limit to prove that the limit f(x)=mc+b as x approaches c?

Answer 1

Choose `delta <= epsilon/abs(m)`.

Explanation:

The if `0 < abs(x-c) < delta`, then we have

`abs(f(x) - (mc+b)) = abs((mx+b)-(mc+b))`

`= abs(mx-mc)`

`= abs(m)abs(x-c)`

`< abs(m)delta` `" "` (see Note below)

`<= abs(m) epsilon/abs(m)`

`= epsilon`

That is:

if `0 < abs(x-c) < delta`, then `abs(f(x) - (mc+c)) < epsilon`.

Therefore, by the definition of limit,

`lim_(xrarrc)(mx+b) = mc+b`

Note : Given `abs(x-c) < delta`, we can multiply on both sides by `absm` -- which is positive -- to get `abs(m)abs(x-c) < abs(m)delta`

Also Note that we must use the most strict inequality between the beginning and the end to connect the first and last expressions.

Final also note: If `m = 0`, then the result is still true, but the proof does not work, because we cannot use `epsilon/absm`. (If `m = 0`, use `delta =` any number you like.)