Question

How do you prove that the limit of `f(x) =2x-3` as x approaches 5 is 7 using the epsilon delta proof?

Answer 1

The function `f(x) = 2x-3` is a polynomial and as such it is continuous for every `x in RR`. Then:

`lim_(x->5) f(x) = f(5) = 2 xx 5 -3 = 7`

To prove it using the limit definition consider the value:

`abs(f(x) - 7) = abs ( 2x-3-7 ) = abs(2x-10) = 2 abs(x-5)`

For `x in (5-delta,5+delta)` with `delta > 0` then we have:

`abs(f(x) - 7) = 2 abs(x-5)< 2delta`

Given any `epsilon > 0` if we then choose `delta_epsilon < epsilon/2` we have that:

`x in (5-delta_epsilon, 5+delta_epsilon) => abs(f(x) - 7) < 2 delta_epsilon < epsilon`

which proves the limit.

Answer 2

Please see below.

Explanation:

Given `epsilon > 0`, let `delta = epsilon/2`.

Now if `0 < abs(x-5) < delta`, then

`abs((2x-3)-7) = abs(2x-10)`

`= 2 abs(x-5)`

`< 2delta` `" "` (since `abs(x-5) < delta`)

# = 2(epsilon/2)

`= epsilon`

That is: if `0 < abs(x-5) < delta`, then `abs((2x-3)-7) < epsilon`