Consider the function

f(x)= 2e^x, if x<0
f(x) = -3x+2, if 0< or equal to x < or equal to 2/3
f(x) = 1, if x > or equal to 2/3

a) evaluate the limit X->0- and X->0+
b) Does limit X ->0 exist? Is f continuous at x=0
c) Evaluate the limit X->2/3- and X -> 2/3+
D) Does limit X->2/3 exist? Is f continuous at x=2/3?

a) To evaluate the limits as X approaches 0 from the left (X->0-) and from the right (X->0+), we need to substitute the values into the function.

For X->0-, we have f(X) = 2e^X. Substituting X = 0, we get f(0) = 2e^0 = 2(1) = 2.

For X->0+, we have f(X) = -3X + 2. Substituting X = 0, we get f(0) = -3(0) + 2 = 2.

b) The limit as X approaches 0 exists because the limit from the left (X->0-) and from the right (X->0+) both equal 2. However, f is not continuous at X = 0 because the left-sided limit (2e^X) and the right-sided limit (-3X + 2) are not equal.

c) To evaluate the limits as X approaches 2/3 from the left (X->2/3-) and from the right (X->2/3+), we substitute the values into the function.

For X->2/3-, we have f(X) = -3X + 2. Substituting X = 2/3, we get f(2/3) = -3(2/3) + 2 = -2 + 2 = 0.

For X->2/3+, we have f(X) = 1. Substituting X = 2/3, we get f(2/3) = 1.

d) The limit as X approaches 2/3 exists because the limit from the left (X->2/3-) and from the right (X->2/3+) both equal 0 and 1 respectively. However, f is not continuous at X = 2/3 because the left-sided limit (-3X + 2) and the right-sided limit (1) are not equal.