A function f(x), a point ( x)0, the limit of f(x) as x approaches (x)0, and a positive number ε is given. Find? a number such that for all x, 0 < < δ ⇒ < ε f(x) =6(x) - 9, L = -3,(x)0 = 1, and ε = 0.01

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Answer 1 · 2017-10-08T09:46:34

The value of #delta=0.00167#

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=0.00167#

#1+delta=1.001666#

#delta=1.001666-1=0.00167#