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

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

1 Answer
Jun 10, 2018

Evaluate the difference:

abs(x^2+3-4) = abs (x^2-1)x2+34=x21

abs(x^2+3-4) = abs (x-1)abs(x+1)x2+34=|x1||x+1|

Let now x=1+xix=1+ξ, then:

abs(x^2+3-4) = abs xi abs ( 2 +xi)x2+34=|ξ||2+ξ|

Given any number epsilon > 0ε>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