F (x) located in the left clinical region of x = X0

F (x) located in the left clinical region of x = X0

First of all, you may be wrong
If f (x) < g (x), LIM (x → x0 -) f (x) = a, LIM (x → x0 -) g (x) = B, then there must be a < B
This conclusion is wrong (the correct conclusion is "a"

Why is the boundedness of F (x) in a centreless neighborhood of x0 a necessary condition for the existence of limf (x), not a sufficient and necessary condition
Why is f (x) unbounded in a centreless neighborhood of x0 a necessary condition for the existence of limf (x) = ∞, not a sufficient and necessary condition

Why f (x) is bounded in a centreless neighborhood of x0 is a necessary condition for the existence of limf (x), not a sufficient and necessary condition
Consider the case that the left and right limits of F (x) are not equal at a certain point!
necessity:
It is defined by limit
∵lim(x→x0)f(x)=∞
For any M > 0, there exists δ > 0, st.0

Given the function f (x) = AX2 + (B-8) x-a-ab, when x ∈ (- 3,2), f (x) > 0, when x ∈ (- ∞, - 3) ∪ (2, + ∞), f (x)

If a is less than 0 and f (x) = 0, the two are - 3 and 2
Then - 3 + 2 = - (B-8) / a (1)
(-3)*2=(-a-ab)/a…… (2)
From (1) and (2), a = - 3, B = 5
Because AX2 + BX + C ≤ 0 holds on [1,4]
That is - 3x ^ 2 + 5x + C ≤ 0 is constant on [1,4]
Then C ≤ 3x ^ 2-5x is constant on [1,4]
Let g (x) = 3x ^ 2-5x
Then G (x) = 3 (X-5 / 6) ^ 2-25 / 12
G (x) decreases monotonically on (- ∞, 5 / 6) and increases monotonically on [5 / 6, + ∞)
Then the minimum value of G (x) on [1,4] is g (1) = - 2
So C ≤ - 2