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#.