How do you prove the statement lim as x approaches 4 for # (x^2) = 4# using the epsilon and delta definition?

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Answer 1 · 2015-10-15T11:17:32

Note that the limit as x approaches 4 is 16, and the limit as x approaches 2 is 4.
I will prove the latter case for you.

By definition, #lim_(x->x_0)f(x)=Liff AA epsilon>0, EE delta >0# such that #|x-x_0| < delta =>|f(x)-L| < epsilon#

Let #epsilon > 0# and select #k in RR^+# such that #|x-2|< k#

#therefore -k < x-2 < k#

#therefore 4-k < x+2 < k+4 #

Hence #|x+2| < k+4#

Now select #delta=min{epsilon/(k+4); k}#

Clearly #delta>0#

Let #|x-2| < delta#

Now : #|f(x)-4|=|x^2-4|#
#=|x-2|*|x+2|#

#< delta*epsilon/delta #

#=epsilon#

This then proves that #lim_(x->2)x^2=4#