AP Calculus BC : Continuity in Terms of Limits

Example Questions

Example Question #1 : Limits And Continuity

The graph above is a sketch of the function . For what intervals is  continuous?

Explanation:

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 .

Example Question #497 : Calculus Ii

Consider the piecewise function:

What is ?

Limit does not exist.

Limit does not exist.

Explanation:

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 .

Example Question #498 : Calculus Ii

Consider the function .

I.

II.

III.

III only

I and II

II only

I and III

I and II

Explanation:

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.