Firstly, recall the rules for evaluating the limits of rational expressions as they go to infinity:

If the degree on the top is greater than the degree on the bottom, the limit does not exist.

If the degree on the top is less than the degree on the bottom, the limit is 0.

If the degrees in both the numerator and denominator are equal, the limit is the ratio of the leading coefficients.

These rules follow from how the function grows as the inputs get larger, ignoring everything but the leading terms.

Then, note that, while the numerator is not written in the correct order for you, that the highest power in the numerator is 5. Likewise, the highest power in the denominator is 5. Thus, the limit will be the ratio of the coefficients on the  terms, which is . 

The equation  will have a horizontal asymptote y=4.

We can find the horizontal asymptote by looking at the terms with the highest power.

The terms with the highest power here are  in the numerator and  in the denominator. These terms will "take over" the function as x approaches infinity. That means the limit will reach the ratio of the two terms.

The ratio is 

First step for finding limits: evaluate the function at the limit.

which is an Indeterminate Form.

This got us nowhere. However, since we have an indeterminate form, we can use L'Hopital's Rule (take the derivative of the top and bottom and the limit's value won't change).

This is something we can evaluate:

The value of this limit is .

The equation given is a model of logistic growth. Note that one of the other common forms this equation might be given in is . What we want to know is what the population will be given unlimited time, or what  is.

Looking at the form of the derivative above, note that if we start with 14 bunnies, we get a positive derivative. This means that the population is increasing. As our function is continuous, the population will keep growing until the derivative hits 0. When does this occur?

In the above form, it should be more clear that the derivative is only 0 at  and . Thus, the population will keep growing, but never go above 350, because if it were to hit 350, the derivative would be 0 and growth would stop. (Alternatively, if the population were above 350, you can see the derivative would be negative and that the population would shrink back down to 350. Indeed, it wasn't necessary to tell you we started with 14 bunnies -- the limit will be the same for any positive starting value.)

Unlike vertical asymptotes, horizontal asymptotes of certain functions may be crossed. In these cases, and in the case of the given function, the horizontal asymptote may be crossed and even have defined values laying on it (e.g., f(0)=0 for the function provided). 

What is meaningful about the horizontal asymptote in this example is that it suggests the behavior of the function at large magnitudes. The denominator will increase much faster than the numerator. That is to say that the  will grow exponentially larger the  in the numerator such that, at large values (both positive and negative), the function will output y-values tending toward y=0.

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.

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   

.

Step 1: Rewrite the problem.

integral (x³+x^1/2) dx from 0 to 1 

Step 2: Integrate 

x⁴/4 + 2x^(2/3)/3 from 0 to 1

Step 3: Plug in bounds and solve. 

[1⁴/4 + 2(1)^(2/3)/3] – [0⁴/4 + 2(0)^(2/3)/3] = (1/4) + (2/3) = (3/12) + (8/12) = 11/12

Integral from 1 to 2 of (1/x³) dx

Integral from 1 to 2 of (x^-3) dx

Integrate the integral. 

 from 1 to 2 of (x^–2/-2)  

(2^–2/–2) – (1^–2/–2) = (–1/8) – (–1/2)=(3/8)