How do you use the epsilon delta definition to find the limit of #(x^2 + 4x) # as x approaches #2#?

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Answer 1 · 2017-07-12T23:47:32

Given any #epsilon > 0#, if we choose #delta_epsilon < min (1,epsilon/9)# then:

#x in (2-delta_epsilon, 2+delta_epsilon) => abs(x^2+4x-12) < epsilon#

so:

#lim_(x->2) x^2+4x = 12#

As #f(x) = x^2 +4x# is a polynomial function, it is continuous for every #x in RR#, then:

#lim_(x->2) f(x) = f(2) = 2^2+4xx2 = 4 + 8 = 12#

Evaluate now:

#abs (f(x) -12) = abs (x^2+4x-12)#

#abs (f(x) -12) = abs ((x-2)(x+6))#

#abs (f(x) -12) = abs (x-2)abs(x+6)#

Given now #epsilon > 0#, choose #delta_epsilon < min(1,epsilon/9)#

For #x in (2-delta_epsilon, 2+delta_epsilon)# we have that:

#abs (x-2) < delta_epsilon < epsilon/9#

And as #(x+6) # is positive in the interval:

#abs(x+6) = x+6 <= 2+delta_epsilon + 6 < 2+1+6 =9#

Then:

#abs (f(x) -12) = abs (x-2)abs(x+6) < epsilon/9 xx 9 = epsilon#