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

2 Answers
Oct 14, 2017

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

Oct 14, 2017

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.