Question

Find a number such that for all x, 0 < < δ ⇒ < ε.???? f(x) = 6x - 9, L = -3, x0 = 1, and ε = 0.01 0.001667 0.01 0.003333 0.000833

Answer 1

The value of `delta=0.001667`

Explanation:

The definition of the limit is

`lim_(x->x_0)f(x)=L`

`AA epsilon>0, EE delta>0` such that

`|f(x)-L|< epsilon`

when `|x- x_0| < delta`

Here,

`f(x)=6x-9`

`x_0=1`

`L=-3`

`epsilon=0.01`

Therefore,

`|6x-9-(-3)|<0.01`

`|6x-6|<0.01`

`-0.01<6x-6<0.01`

`-0.01+6<6x<0.01+6`

`5.99/6< x < 6.01/6`

`0.998333< x <1.001666`
Also,

`|x-1| < delta`

`-delta<x-1 < delta`

`1-delta< x < 1+delta`

Therefore,

`1-delta=0.998333`

`delta=1-0.998333=0.001667`

`1+delta=1.001666`

`delta=1.001666-1=0.001667`

Answer 2

Also, for this example, given any `epsilon>0`, the value `delta=epsilon/6>0` will be such that `|f(x) - L| < epsilon` for all `x` satisfying `|x - x_0| < delta`.

Explanation:

If `|x - x_0| = |x - 1| < delta = epsilon/6`, then

`|f(x) - L| = |6x-9+3|`

`=|6(x-1)|=6|x-1| < 6 * epsilon/6 = epsilon`.

This proves that `lim_{x->1}(6x-9)=-3`.