Given any ϵ>0, we need to find a number δ>0

Question

Let f(x)=x^2−8x. To prove that limx→4f(x)=−16, we proceed as follows. Given any ϵ>0, we need to find a number δ>0 such that if 0<|x−4|<δ, then |(x2−8x)−(−16)|<ϵ. What is the (largest) choice of δ that is certain to work? (Your answer will involve ϵ. When entering your answer, type e in place of ϵ.)

δ=_______

Answer 1

Not a fan of real analysis, hence never excelled in it...

was mediocre at best....

but it goes something like this...going backwards....

 

| x^2 - 8x - (-16) | < e

| x^2 - 8x + 16 | < e

| (x-4)^2| < e    

 (x-4)^2 < e  <---- squaring a number guarantees positive result; absolute values dropped

 

 x-4 < square-root(e)

 

 0< x < square-root(e) +4   <---- there's your delta

 

So now going the other way,  e>0 --> delta = square-root(e)+4 > 0

 

 so if x is in a neighborhood of delta, then |X|<delta

 

  -delta < x < delta

 

-4 -delta < x-4 < delta-4    <-- subtracts 4 from all 3 sides

 

 -(delta-4) < x-4 < delta-4

 

0< | X-4 | < delta-4     

 

| X-4 |^2 < (delta-4)^2

 

|X^2 - 8x + 16| < (delta-4)^2

 

|X^2 - 8x - -16| < (delta-4)^2

 

| f(x) - L | < (delta -4)^2

 

so there's the function being in the proper neighborhood on the left side. RIght side had better collapse to e.

 

 (delta-4)^2 = delta^2 - 8*delta + 16 

                    = [ square-root(e)+4]^2 - 8*[square-root(e)+4] + 16

 

                    =  e + 8(square-root(e) + 16 - 8*square-root(e) -32 + 16

 

                    = e  

 

yep! the only issue I can see here is the step I highlighted in bold. That is the square of an absolute value

  is the absolute value of the square and vice versa. As long as everything is positive. it should be ok.

  May wanna double check that, but there's a good start for you at least. good luck