Please answer these with lots of detail. I want to be able to understand the topic and material well.

Step-by-step explanation:

Q3

(a) If 0 < |x − 1| < δ, then |f(x) − 2| < 2

What this means is, how far can x stray from the x=1 line such that f(x) stays within 2 units of the y=2 line (0 < f(x) < 4).

If we move 2 units left of x=1, we get f(x) = 4.

If we move about 3.5 units right of x=1, we get f(x) = 4.

Therefore, δ can't be more than 2.

(b) If 0 < |x| < δ, then |f(x) − 3| < 1

What this means is, how far can x stray from the x=0 line such that f(x) stays within 1 unit of the y=3 line (2 < f(x) < 4).

If we move 1 unit left of x=0, we get f(x) = 4.

If we move 1 unts right of x=0, we get f(x) = 2.

Therefore, δ can't be more than 1.

Since f(x) isn't continuous within this domain, we can't conclude that the limit exists.

Q4

(a) Yes.  If δ = 0.25, then 0.75 < x < 1.25, and f(x) > 200.

(b) No.  f(1) = 300, so even if δ = 0, f(x) will be less than 400.

(c) Yes.  If δ ≈ 0.1, then 0.9 < x < 1, and f(x) > 450.

75 tickets have been sold so far.

step-by-step explanation:

in this question it is clearly mentioned that 45 children's tickets have been sold and this number is 60% of the total tickets sold.

now let's assume the total number of tickets sold be x.

then 60% of x = 45

⇒

⇒ x(.6) = 45

by dividing with .6 on both the sides of the equation

⇒ x = 75

therefore 75 tickets have been sold so far.