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

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

1 Answer
Dec 22, 2017

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.