Need help with these 2 questions please and thanks

These are two questions and two answers.

1) Problem # 15.

(i) [tex] \lim_{x \to \ 60^{-} } f(x) [/tex]

You have to approach the value of x by the left. That is the horizontal segment at y = 56 (with the open circle to the left and the solid circle to the right).

So, the value of the limit es 56.

(ii) [tex] \lim_{x \to \ 60^{+} } f(x)[/tex]

You have to approach the x value by the right. That is the horizontal segment  at y = 68.

So the value of the limit is 68.

(iii) Since, the existence of the limit requires that both limits from the left and the right be equal, the conclusion is that the limit does not exist.

That is shown in the graph because the way the function is defined is different if the value of x is greater or less than 60.

(iv) What causes the graph to jump vertically by the same amount at discontinuites is that the rates are defined in intervals and not in a continuous way.

2) Problem 16. Mathematical induction.

Mathematical induction requires 3 steps: 1) Initial hypothesis, 2) assume the equation is valid for n = k, and 3) prove the equation is valid for n = k+1.

This is the solution step by step.

1) Initial hypothesis: first term ⇒ n = 1

left side: 8
rigth side: 4n(n+1) = 4(1+1) = 4(2) = 8

8 = 8 ⇒ check

2) Assume n = k

letf side: 8 + 16 + 24 + ... + 8k
right side: 4k(k + 1)

3) Proove for n = k + 1

left side: 8 + 16 + 24 + ... + 8k + 8(k+1)
right side: 4(k+1) (k + 1 + 1) = 4(k+1)(k+2)

Since 8 + 16 + 24 + ... 8k is assumed to be equal to 4k(k+1), then the left side ends as:

left side: 4k(k+1) + 8(k+1)

take common factor k + 1 → (k+1)(4k + 8)
take common factor 4 for the second parenthesis → (k+1) 4(k + 2) = 4(k+1)(k+2)

So, we have shown that the left side is equal fo the right side, which completes the proof by induction.