Continuity in Terms of Limits - AP Calculus BC

For a function to be continuous at a point , must exist and .

This is true for all values of except and .

Therefore, the interval of continuity is .

For a function to be continuous at a particular point, the limit of the function at that point must be equal to the value of the function at that point.

First, notice that

.

This means that the function is continuous everywhere.

Next, we must compute the limit. Factor and simplify f(x) to help with the calculation of the limit.

Thus, the limit as x approaches three exists and is equal to , so I and II are true statements.

For a function to be continuous at a point , must exist and .

This is true for all values of except and .

Therefore, the interval of continuity is .

For a function to be continuous at a particular point, the limit of the function at that point must be equal to the value of the function at that point.

First, notice that

.

This means that the function is continuous everywhere.

Next, we must compute the limit. Factor and simplify f(x) to help with the calculation of the limit.

Thus, the limit as x approaches three exists and is equal to , so I and II are true statements.

The piecewise function

indicates that is one when is less than five, and is zero if the variable is greater than five. At , there is a hole at the end of the split.

The limit does not indicate whether we want to find the limit from the left or right, which means that it is necessary to check the limit from the left and right. From the left to right, the limit approaches 1 as approaches negative five. From the right, the limit approaches zero as approaches negative five.

Since the limits do not coincide, the limit does not exist for .

For a function to be continuous at a point , must exist and .

This is true for all values of except and .

Therefore, the interval of continuity is .

For a function to be continuous at a particular point, the limit of the function at that point must be equal to the value of the function at that point.

First, notice that

.

This means that the function is continuous everywhere.

Next, we must compute the limit. Factor and simplify f(x) to help with the calculation of the limit.

Thus, the limit as x approaches three exists and is equal to , so I and II are true statements.

The piecewise function

indicates that is one when is less than five, and is zero if the variable is greater than five. At , there is a hole at the end of the split.

The limit does not indicate whether we want to find the limit from the left or right, which means that it is necessary to check the limit from the left and right. From the left to right, the limit approaches 1 as approaches negative five. From the right, the limit approaches zero as approaches negative five.

Since the limits do not coincide, the limit does not exist for .

The piecewise function

indicates that is one when is less than five, and is zero if the variable is greater than five. At , there is a hole at the end of the split.

The limit does not indicate whether we want to find the limit from the left or right, which means that it is necessary to check the limit from the left and right. From the left to right, the limit approaches 1 as approaches negative five. From the right, the limit approaches zero as approaches negative five.

Since the limits do not coincide, the limit does not exist for .

For a function to be continuous at a point , must exist and .

This is true for all values of except and .

Therefore, the interval of continuity is .

For a function to be continuous at a particular point, the limit of the function at that point must be equal to the value of the function at that point.

First, notice that

.

This means that the function is continuous everywhere.

Next, we must compute the limit. Factor and simplify f(x) to help with the calculation of the limit.

Thus, the limit as x approaches three exists and is equal to , so I and II are true statements.

For a function to be continuous at a point , must exist and .

This is true for all values of except and .

Therefore, the interval of continuity is .

For a function to be continuous at a particular point, the limit of the function at that point must be equal to the value of the function at that point.

First, notice that

.

This means that the function is continuous everywhere.

Next, we must compute the limit. Factor and simplify f(x) to help with the calculation of the limit.

Thus, the limit as x approaches three exists and is equal to , so I and II are true statements.

The piecewise function

indicates that is one when is less than five, and is zero if the variable is greater than five. At , there is a hole at the end of the split.

The limit does not indicate whether we want to find the limit from the left or right, which means that it is necessary to check the limit from the left and right. From the left to right, the limit approaches 1 as approaches negative five. From the right, the limit approaches zero as approaches negative five.

Since the limits do not coincide, the limit does not exist for .

For a function to be continuous at a point , must exist and .

This is true for all values of except and .

Therefore, the interval of continuity is .

For a function to be continuous at a particular point, the limit of the function at that point must be equal to the value of the function at that point.

First, notice that

.

This means that the function is continuous everywhere.

Next, we must compute the limit. Factor and simplify f(x) to help with the calculation of the limit.

Thus, the limit as x approaches three exists and is equal to , so I and II are true statements.

The piecewise function

indicates that is one when is less than five, and is zero if the variable is greater than five. At , there is a hole at the end of the split.

The limit does not indicate whether we want to find the limit from the left or right, which means that it is necessary to check the limit from the left and right. From the left to right, the limit approaches 1 as approaches negative five. From the right, the limit approaches zero as approaches negative five.

Since the limits do not coincide, the limit does not exist for .

For a function to be continuous at a point , must exist and .

This is true for all values of except and .

Therefore, the interval of continuity is .

For a function to be continuous at a particular point, the limit of the function at that point must be equal to the value of the function at that point.

First, notice that

.

This means that the function is continuous everywhere.

Next, we must compute the limit. Factor and simplify f(x) to help with the calculation of the limit.

Thus, the limit as x approaches three exists and is equal to , so I and II are true statements.