How do you prove a limit of (x^2+3) as x approaches 1 equal 4? Thanks

Question

Please show the process using Epsilon-Delta proof. Thanks again

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Answer 1 · 2018-06-10T21:26:16

Evaluate the difference:

#abs(x^2+3-4) = abs (x^2-1)#

#abs(x^2+3-4) = abs (x-1)abs(x+1)#

Let now #x=1+xi#, then:

#abs(x^2+3-4) = abs xi abs ( 2 +xi)#

Given any number #epsilon > 0# choose:

#delta_epsilon < min(1,epsilon/3)#

For #x in (1-delta_epsilon,1+delta_epsilon)# we have that:

#abs xi < delta_epsilon#

Now, as #delta_epsilon < 1# we have that:

#abs(2+xi) <= 2+abs xi < 2+1 = 3#

and then:

#abs(x^2+3-4) = abs xi abs ( 2 +xi) < 3abs(xi)#

and because #delta_epsilon < epsilon/3# we have that:

#abs(x^2+3-4) < 3abs(xi) < 3 * epsilon/3 = epsilon#

We can conclude that for every #epsilon# if we choose #delta_epsilon < min(1,epsilon/3)# we have that:

#x in (1-delta_epsilon,1+delta_epsilon) => abs(x^2+3-4) < epsilon#