Lim x → 3 − f ( x ) = 6 and lim x → 3 + f ( x ) = 5 . As x approaches 3 , f ( x ) approaches 5 , but f ( 3 ) = 6 . As x approaches 3 from the right, f ( x ) approaches 6 . As x approaches 3 from the left, f ( x ) approaches 5 . As x approaches 3 , f ( x ) approaches 6 , but f ( 3 ) = 5 . As x approaches 3 from the left, f ( x ) approaches 6 . As x approaches 3 from the right, f ( x ) approaches 5 . In this situation is it possible that lim x → 3 f ( x ) exists? Explain.

Step-by-step explanation:

In limits and continuity of a function, for a function to be continuous, the left hand limit of the function must be equal to the left hand limit and they must be equal to the limit of the function at that point. Mathematically;

All continuous functions are known to exist but discontinuous functions doesn't exist.

From the question, the point we are considering is at n =3

Given

but f(3)= 6

As we can see, the left hand limit of the function isn't equal to the left hand limit, this shows that the limit of the function isn't continuous and since all discontinuous function does not exists then  lim x → 3 f ( x ) does not exist.

dc=6 square root of 3 and ac is 12 square root 3

c and d

step-by-step explanation:

square root of 3/2 = 18/c

cross multiply to get 18*2=√3*c

36/√3 * √3/√3 = c

c= 36*√3/3

c=12√3