Question

Is the given piecewise function continuous at x=0 ? f(x)={(x²+3,x!=0),(4,x=0):}

Answer 1

The piecewise function f(x) is not continuous at x=0 because the limit as x approaches 0 is 3, which does not equal the defined value of f(0) at 4.

Step-by-step explanation:

To determine if the given piecewise function f(x) is continuous at x=0, we need to check if the following conditions are met:

• The function is defined at x=0.
• The limit of f(x) as x approaches 0 exists.
• The limit of f(x) as x approaches 0 from both sides is equal to f(0).

The function is given by f(x) = { (x² + 3, x != 0), (4, x = 0) }. It is defined at x=0 since the value f(0)=4 is provided. To find the limit as x approaches 0, we consider the x² + 3 portion of the function, which gives limx→0(x² + 3) = 0² + 3 = 3. This limit does not equal f(0), which means the function is not continuous at x=0.

The complete question is: Is the given piecewise function continuous at x=0 ? f(x)={(x²+3,x!=0),(4,x=0):} is: