How do you prove that #lim_(x->5)x^2!=24# using limit definition?

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Answer 1 · 2017-12-22T02:31:53

Pose #x=xi+5# and consider the quantity:

#abs(x^2-24) = abs((xi+5)^2-24)#

#abs(x^2-24) = abs(xi^2+10xi+25-24)#

#abs(x^2-24) = abs(xi^2+10xi+1)#

Using the reverse triangular inequality we can state that:

#(1) " "abs(x^2-24) >= abs(1-abs(xi^2+10xi))#

Consider now any #delta >0#.
We have that for #x in (5-delta, 5+delta)# then #abs xi < delta #,

and based on the triangular inequality:

#abs(xi^2+10xi) <= xi^2+10abs xi < delta^2+10delta#

If we choose #delta < 1/22< 1# then #delta^2 < delta# and:

#abs(xi^2+10xi) < 11 delta < 11*1/22 = 1/2 #

so that based on #(1)#:

#abs(x^2-24) >= abs(1-1/2) = 1/2#

So if we choose any #epsilon# such that #0 < epsilon < 1/2# we can state that for #x in (5-1/22,5+1/22)# we have that:

#abs(x^2-24) >= epsilon#

which contradicts the limit definition.