Limits - Pre-Calculus

Since the factor is in the numerator and the denominator, there is a removable discontinuity at . The function is not defined at , but function would move towards the same point for the resultant function .

Since the factor cannot be factored out, there is an infinite discontinuity at . The denominator will get very small and the numerator will move toward a fixed value.

There is no discontinuity at all at . The function simply evaluates to zero at this point.

The function in the numerator factors to:

so if we cancel the x+3 in the numerator and denominator we have the same function but it is continuous. The gives us a hole at x=-3 so our function is not continuous at x=-3.

In order to determine if a function is continuous at a point three things must happen.

1. Taking the limit from the lefthand side of the function towards a specific point exists.

2. Taking the limit from the righthand side of the function towards a specific point exists.

3. The limits from 1) and 2) are equal and equal the value of the original function at the specific point in question.

In our case,

Because all of these conditions are met, the function is continuous at 0.

First, find that at any point where , .

Then find that

and .

As these are all equal, it can be determined that the function is continuous on all points of its domain.

As approaches , the function approaches , which is undefined. However, if we factor , we get:

The factors in the numerator and denominator cancel out, leaving .

Therefore, our function is continuous at all values of from .

This is a graph of . We know that is undefined; therefore, there is no value for . But as we take a look at the graph, we can see that as approaches 0 from the left, approaches negative infinity.

This can be illustrated by thinking of small negative numbers.

NOTE: Pay attention to one-sided limit specifications, as it is easy to pick the wrong answer choice if you're not careful.

is actually infinity, not negative infinity.

The function given is a polynomial with a term , such that is greater than 1.

Whenever this is the case, we can say that the whole function diverges (approaches infinity) in the limit as approaches infinity.

This tells us that the given function is not a very realistic description of a car's speed for large !

will approach when approaches , so will be of type as shown below:

So, we can apply the L’ Hospital's Rule:

since:

hence:

This can be rewritten as follows:

We can substitute , noting that as , :

, which is the correct choice.

The function has a removable discontinuity at . Once a factor of is "divided out" the resultant function is , which evaluates to as approaches 0.

As x approaches infinity we only need to look at the highest order of polynomial in both the numerator and denominator. Then we compare the highest order polynomial in both the numerator and denominator. If the denominator is higher order our limit goes to zero, if the numerator is higher our order our limit goes to positive or negative infinity (depending on the sign of the highest order x term). If our numerator and denominator have the same order the limit goes to a/b where a is the coefficient for the highest order x in the numerator and b is the coefficient for the highest order x in the denominator.

Our numerator has higher order and the coefficient for the x to the fourth term is negative so our limit goes to negative infinity.

As x approaches infinity we only need to look at the highest order of polynomial in both the numerator and denominator. Then we compare the highest order polynomial in both the numerator and denominator. If the denominator is higher order our limit goes to zero, if the numerator is higher our order our limit goes to positive or negative infinity (depending on the sign of the highest order x term). If our numerator and denominator have the same order the limit goes to a/b where a is the coefficient for the highest order x in the numerator and b is the coefficient for the highest order x in the denominator.

For our equation the orders are the same in x for the numerator and denominator (both 4). So we divide the coefficients of the highest order x terms to get our limit and we get

When x=3/2 our denominator is zero so we can't just plug in 3/2 to get our limit. If we look at the numerator when x=3/2 we find that it is zero as well so our numerator can be factored. We see that our limit can be re-written as:

we then can cancel the 2x-3 from the numerator and denominator leaving us with:

and we can just plug in 3/2 into this limit to get

note: our function is not continuous at x=3/2 but the limit does exist.

To solve this problem we need to expand the term in the numerator

when we do that we get

the second degree x terms cancel and we get

now we can cancel our h's in the numerator and denominator to get

then we can just plug 0 in for h and we get our answer

First factor the numerator to simplify the function.

,

so

.

Now

.

There is no denominator now, and hence no discontinuity. The limit can be found by simply plugging in for .

.

There are a number of ways to solve this. Either we can solve by finding what K(C) evaluates to for larger values of C and see where they converge, i.e. :

K(10000) = 952.38

K(1000000) = 999.5

etc...

And we see that K(C) approaches 1000 for larger concentrations, C.

Or we can notice that the dominant term in the numerator is 1000C; dominant term in the denominator is C.

1000C / C = 1000, which will ultimately be our limit.

To find here, you need only plug in for :

Find a common denominator for both the upper and lower expressions and then simplify:

Substituting the value of will yield , which is not in indeterminate form. Therefore, L'Hopital's rule cannot be used.

The question asks for the limit as x approaches to five from the right side in the graph. As the graph approaches closer and closer to , the y-value decreases to negative infinity and will never touch .

The correct answer is:

In the unsimplified form, the limit does not exist; however, the numerator can be factored and simplified.