Calc

Introduction:

Let g(x) = x2 – x – 7 be a function from the real numbers to the real numbers.

Task:

A. Find the set of δ values that satisfy the formal definition of limx4 g(x) = 5 when given

the value ε = 0.3, showing all work.

0 < |g(x)-5| < 0.3 0 < |(x^2 -x-7) -5| < 0.3 4.7 < (x^2 -x -7 )< 5.3

4.7 = x^2 -x -7 0 = x^2 -x -11.7 x = 3.957 or x = -2.957

x^2 -x -7 = 5.3 0 = x^2 -x -12.3 x = 5.35 or x = 4.03

0 < |x-4| < δ ; x < δ+4 ; 3.968 < x< 4.03; δ = .031 -3.04 < x < -2.956877; δ = .085 (however this value would not apply to x approaching 4)

B. Use the formal definition of a limit to demonstrate that limx4 g(x) = 5, showing all

work.

We can see that 4^2 -4 -7 =5 however we can also check using δ notation.

Using the work completed above we can see:

4.7 < (x^2 -x -7 )< 5.3 4.7< ((4-δ)^2 -(4- δ) -7)) 4.7 < ((4-.031)^2 -(4-.031) -7 4.7 < 4.783

((4+δ)^2 -(4+ δ) -7)) < 5.3 ((4 +.031)^2 -(4+.031) -7 5 and ε < 5 . We will start with ε > 5, so first let's say ε = 5. This gives us a domain to map of (0,10) We will now want a delta such that (5 + δ )^2 =10 so let's choose δ =√10 -5 . This is defined on the interval (√10,10 -√10) this gives us approximately (3.162, 6.8377). This will then map to { (√10)^2-√10-7, (10-√10)^2- (10-√10)-7}in our range and is contained in (0,10).

Now for ε > 5, we can widen the interval in the desired range, but the interval in the domain stays the same and is mapped to the same sub-interval in our range. This means δ =√10 -5 works for any ε > 5.

Now suppose 0 < ε < 5 and want to find a δ such that 0< |(x^2-x-7) -5| < ε , when 0 0 .

We now choose the smaller of the two extremes for our interval. This gives us (ε+ 5)-4 > ( 5-ε) -4 for 0< ε < 5, so we use δ =(ε +5)-4 . We can now try with ε =.02. This gives us δ = .0026

4.98< (x^2 -x -7 )< 5.02 4.98< ((4-δ)^2 -(4- δ) -7)) 4.98 < ((4-.0026)^2 -(4-.0026) -7 4.98...