Good answers get brainliest. 9.04 1. find the indicated limit, if it exists. limit of f of x as x approaches 8 where f of x equals x plus 10 when x is less than 8 and f of x equals 10 minus x when x is greater than or equal to 8 a) 8 b) 18 c) the limit does not exist. d) 2 2. find the indicated limit, if it exists. limit of f of x as x approaches 5 where f of x equals 5 minus x when x is less than 5, 8 when x equals 5, and x plus 3 when x is greater than 5 a) 0 b) 8 c) 3 d) the limit does not exist. 3. find the indicated limit, if it exists. limit of f of x as x approaches 0 where f of x equals 5 x minus 9 when x is less than 0 and the absolute value of the quantity 2 minus x when x is greater than or equal to 0 a) -9 b) -7 c) 2 d) the limit does not exist. 4. use graphs and tables to find the limit and identify any vertical asymptotes of limit of 1 divided by the quantity x minus 4 as x approaches 4 from the left . (2 points) a) ∞ ; x = -4 b) -∞ ; x = -4 c) -∞ ; x = 4 d) 1 ; no vertical asymptotes 5. give an example of a function with both a removable and a non-removable discontinuity.

1. We analyze the limit by approaching it from both the left and the right.
From the left: f(x) = x + 10 (for x < 8), as x --> 8, f(x) --> 18
From the right: f(x) = 10 - x (for x >= 8), as x --> 8, f(x) --> 2
Since the limits on either side do not converge to the same point, the limit does not exist (this is choice C).

2. Using a similar approaching as in #1:
From the left: f(x) = 5 - x (for x < 5), as x --> 5, f(x) --> 0
At x = 5 itself: f(x) = 8
From the right: f(x) = x + 3 (for x > 5), as x --> 5, f(x) --> 8
Although the value at x = 5 matches with the limit when approaching from the right, the limit when approaching from the left doesn't match, so the limit does not exist (choice D).

3. From the left: f(x) = 5x - 9 (for x < 0), as x --> 0, f(x) --> -9
From the right: f(x) = |2 - x| (for x >= 0), as x --> 0, f(x) --> 2
Again, since the limits when approaching from the left and right don't match, the limit does not exist. (This is Choice D).

4. lim 1/(x - 4) as x -->4-
If we are approaching x = 4 from the left, we can test values such as 3, 3.9, 3.99, 3.999, approaching 4. For x = 3, f(x) = -1. For x = 3.9, f(x) = -10. For x = 3.99, f(x) = -100. For x = 3.999, f(x) = -1000. This shows that the value continues to go towards negative infinity.
If we were to graph these 4 points on the Cartesian plane, it would also show a curve to slopes downwards to negative infinity, with the vertical asymptote at x = 4. The correct answer is Choice C) -∞ ; x = 4.

5. f(x) = (x+1)(x-1) / [(x+1)(x-2)] is an example of a function with both a removable and non-removable discontinuity.
In this case, because x+1 cancels out from the numerator and denominator, it results in a hollow or missing point (removable) discontinuity at x = -1. This means that the limit still exists as x --> -1. On the other hand, x = 2 is a non-removable discontinuity, since it cannot be cancelled out, and it will be an asymptote.

Yeah she/ he needs look how long that is man