A piecewise function is given. Use the function to find the indicated limits, or state that a limit does not exist.

(a) lim is over x-->d – f(x), (b) lim is over x-->d + f(x), and (c) lim is over x-->d f(x)

x^2 - 5 if x < 0
f(x)= -2 if x ->0 ; d = -3

1. (a)-5 (b)-2 (c)does not exist
2. (a)4 (b)4 (c)4
3. (a)-2 (b)-5 (c)does not exist
4. (a)-5 (b)-2 (c)-2

what is wrong with answer 2? Have you worked this? Iwill be happy to critique your thinking.

To find the limits indicated, we need to evaluate the function for each case.

(a) To find lim as x approaches d from the left, we need to evaluate f(x) when x approaches d from values less than d.

In this case, d = -3. Since the given piecewise function is defined only for x < 0, and d = -3 which is greater than 0, there are no x values less than d.

Therefore, lim as x --> d- f(x) does not exist.

(b) To find lim as x approaches d from the right, we need to evaluate f(x) when x approaches d from values greater than d.

In this case, d = -3. Since the given piecewise function is defined for x = -2 when x approaches 0 from the right, we can evaluate f(x) as -2.

Therefore, lim as x --> d+ f(x) = -2.

(c) To find lim as x approaches d, we need to evaluate f(x) when x approaches d from both the left and the right.

In this case, both lim as x --> d- f(x) and lim as x --> d+ f(x) are defined.

lim as x --> d- f(x) is not defined since there are no x values less than d.

lim as x --> d+ f(x) = -2 since the function evaluates to -2 at x = -2.

Therefore, lim as x --> d f(x) does not exist.

So the correct answer would be option 3. (a) -2 (b) -5 (c) does not exist.