Functions, Graphs, and Limits - AP Calculus AB

The answer is both.

If graphed the student will see that the two graphs are continuous at . There is no gap in the graph or no uneven transitions. If the graph is continuous then it is differentiable so it must be both.

A removeable discontinuity occurs whenever there is a hole in a graph that could be fixed (or "removed") by filling in a single point. Put another way, if there is a removeable discontinuity at , then the limit as approaches exists, but the value of does not.

For example, the function contains a removeable discontinuity at . Notice that we could simplify as follows:

, where .

Thus, we could say that .

As we can see, the limit of exists at , even though is undefined.

What this means is that will look just like the parabola with the equation EXCEPT when, where there will be a hole in the graph. However, if we were to just define , then we could essentially "remove" this discontinuity. Therefore, we can say that there is a removeable discontinuty at .

The functions

, and

have discontinuities, but these discontinuities occur as vertical asymptotes, not holes, and thus are not considered removeable.

The functions

and are continuous over all the real values of ; they have no discontinuities of any kind.

The answer is

.

The mid-point Reimann sum is given by this formula:

, where , and is the number of

intervals. Thus, if the region from to is divided into twenty intervals, and . For ten intervals, and . For five internals, and . The higher the number of intervals, the more precise the estimation. Thus, when (and hence ), the estimation is the most accurate.

To check if the piecewise function is continuous, all we need to do is check that the values at 3 and 5 line up.

At , this means checking that and have the same value. More formally, we are checking to see that , as to be continuous at a point, a function's left and right limits must both equal the function value at that point.

Plugging 3 into both, we see that both of them are 12 at . Thus, they meet up smoothly.

Next, for , we have and . Plugging in 5, we get 22 for both equations.

As all three equations are polynomials, we know they will be continuous everywhere else, and they meet up smoothly at the piecewise bounds, thus ensuring that the function is continuous everywhere.

Note, there are sharp turns at and , but this only means the function isn't differentiable at these points -- we're only concerned with continuity, which is if the equations meet up. Thus, the function is continuous.

is not differentiable at and because the values are discontinuities. is not differentiable at because that point is a corner, indicating that the one-side limits at are different. is differentiable:the one side limits are the same and the point is continuous.

Note that , indicating that there is a horizontal tangent on at . More specifically, the derivative is the slope of the tangent line. If the slope of the tangent line is 0, then the tangent is horizontal.

The other two are incorrect because sharp turns only apply when we want to take the derivative of something. The derivative of a function at a sharp turn is undefined, meaning the graph of the derivative will be discontinuous at the sharp turn. (To see why, ask yourself if the slope at is positive 1 or negative 1?) On the other hand, integration is less picky than differentiation: We do not need a smooth function to take an integral.

In this case, to get from to , we took an integral, so it didn't matter that there was a sharp turn at the specified point. Thus, neither function had any discontinuities.

Unless we are explicitly told so, via graph, information, or otherwise, we cannot assume is continuous at unless , which is required for to be continuous at .

We cannot assume anything about the existence of , because we do not know what is, or its end behavior.

For a function to be continuous the following criteria must be met:

1. The function must exist at the point (no division by zero, asymptotic behavior, negative logs, or negative radicals).
2. The limit must exist.
3. The point must equal the limit. (Symbolically, ).

It is easiest to first find any points where the function is undefined. Since our function involves a fraction and a natural log, we must find all points in the domain such that the natural log is less than or equal to zero, or points where the denominator is equal to zero.

To find the values that cause the natural log to be negative we set

Therefore, those x values will yield our points of discontinuity. Normally, we would find values where the natural log is negative; however, for all the function is positive.

You can substitute to write this as:

Note that as ,

, since the fraction becomes indeterminate, we need to take the derivative of both the top and bottom of the fraction.

, which is the correct choice.

Substitute to rewrite this limit in terms of u instead of x. Multiply the top and bottom of the fraction by 2 in order to make this substitution:

(Note that as , .)

, so

, which is therefore the correct answer choice.

Factor the numerator and simplify the expression.

Evaluate the function at x=2.

There is a discontinuity at x=2, but since it the limit as x approaches 2 from the right is equal to the limit as x approaches 2 from the left, the limit exists.

Factor the numerator to evaluate the limit:

Evaluate the limit:

There is a discontinuity at x=0 but the limit is equal to 8 because the limit from the right is equal to the limit from the left.

As x approaches infinity, the x values that are raised to the highest power dominate the function and all other values become negligible. In this limit, x3 has the highest order in the numerator and 5x2 has the highest power of all the values in the denominator. We can simplify the limit to the following:

After factoring out x2 it is easy to see that the limit diverges to infinity.

First we notice that substituting 5 in for x will give us a 0 in the denominator.

So we simplify the equation by noticing the numerator is the difference of two squares.

Now we can substitute 5 in for x, and we arrive at our answer of 10.

The first step to evaluating limits is always attempt to direct substitute the value into the function. In this case, that is all you need to do.

To start this problem, we need to factor the quadratic equation. This will result in a hole at , which will allow us to directly substitute the limit value into the resulting function to determine the limit.

To evaluate this limit, we need to direct substitute the value in question into our function. Note, this value is in the domain of the function, so direct substitution is the only method we need here.

This limit can't be evaluated by a simple substitution; is not defined at , so some simplification is in order first.

. (Start)

. (Factor the numerator)

. (Cancel the terms)

. (Substitute in the )

.