Limits of Functions (including one-sided limits) - AP Calculus AB

The limit of a function as approaches a value exists if and only if the limit from the left is equal to the limit from the right; the actual value of is irrelevant. Since the function is piecewise-defined, we can determine whether these limits are equal by finding the limits of the individual expressions. These are both polynomials, so the limits can be calculated using straightforward substitution:

, so .

Factor x-4 out of the numerator and simplify:

Evaluate the limit for x=4:

Although there is a discontinuity at x=4, the limit at x=4 is 10 because the function approaches ten from the left and right side.

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.

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 )

.