## Tuesday, June 30, 2015

### Advanced Math Solutions – Limits Calculator, Rational Functions

In the previous post, we learned how to find the limit of a function with a square root in it. Today we will be dealing with rational functions, which are handled differently than the function in the previous posts because rational functions follow a set of rules. A rational function is a function with a ratio of two polynomials, F(x)=\frac{P(x)}{Q(x)}.

Here are the rules:
1. If the degree of P(x) is less than the degree of Q(x), then the limit is 0.
2. If the degree of P(x) and Q(x) are the same, then divide the coefficients of the terms with the largest exponent to get the limit.
3. If the degree of P(x) is greater than the degree of Q(x), then the limit is either \infty or -\infty depending on the signs of polynomials. If both the coefficients of the largest exponent are either negative or both positive, then the limit is \infty. Otherwise, the limit is -\infty.

Here is an example of Rule 1 (click here):

Here is an example of Rule 2 (click here):

Here is an example of Rule 3 (click here):

Rational functions are easy to solve as long as you remember those three rules. Next post, we will talk about L’Hopital’s rule.

Until next time,
Leah

## Tuesday, June 23, 2015

### Advanced Math Solutions – Limits Calculator, Functions with Square Roots

In the previous post, we talked about using factoring to simplify a function and find the limit. Now, things get a little trickier. We can run into the same problem of not being able to initially use substitution when given a function with square roots. Keep in mind the identity (a+b)(a-b)=a^2-b^2. We multiply the numerator and the denominator by the conjugate in order to eliminate the square root. It is important to remember to not immediately multiply out everything. There is a chance that you can use factoring after rationalizing the function. Afterwards, the function looks a lot more attractive and you can plug and chug the x value to get the limit.

Here we will work out the first problem step by step (click here):

1. Try Substitution
By substituting in 4, we get that the limit approaches an indeterminate form. Therefore, this method does not work.

2. Rationalize the function by multiply the numerator and denominator by the conjugate
\quad\quad\quad\lim_{x\to4}\left(\frac{\sqrt{x}-2}{x-4}\cdot\frac{\sqrt{x}+2}{\sqrt{x}+2}\right)
\quad\quad\quad\lim_{x\to4}\left(\frac{x-4}{(x-4)(\sqrt{x}+2)}\right)

We multiplied the numerator and denominator by the conjugate of √x-2 in order to get rid of the radical. Make sure not to multiply out the denominator in case we can factor out something.

3. Factoring
We factored out (x-4) because it was in the numerator and the denominator.

4. Substitution

Here’s another example (click here):

In this next example we do not need to multiply the numerator and denominator by the conjugate. We are able to algebraically manipulate the function in order to rationalize it.

Here’s the example (click here):

There are most steps to this method, which may make it seem hard. As long as you write out each step and pay attention to the sign changes, you should be able to do it. Next post will be about solving limits of rational functions.

Until next time,

Leah

## Monday, June 15, 2015

### Advanced Math Solutions – Limits Calculator, Factoring

In a previous post, we talked about using substitution to find the limit of a function. Sometimes substitution does not immediately work, like when the limit is one of the indeterminate forms. In this case, we must algebraically manipulate the function first. We will focus on factoring the function in order to use substitution.

Here we will work out the first problem step by step (click here):

1. Try Substitution:
When we substitute 2 in for x, we get  0/0, an indeterminate. Therefore, we must manipulate the fraction.

2. Factor both the numerator and denominator:

Once factored, the numerator and the denominator showed that they have a common factor, (x-2). This factor was then cancelled out, leaving a simplified fraction.

3. Substitute

Here is another example using factoring (click here):

Last example … (click here):

I did not find factoring to be too challenging. The only obstacle I came across when dealing with limits was remembering that factoring is an option. Don’t forget that there are multiple ways to solve the limit.

Until Next time,

Leah

Hey Guys! I’m Leah Weil. I am interning with Symbolab for the summer and will be writing some of the upcoming blogs. I will tell you a little bit about myself. I am from Cincinnati, OH, but I currently live in Columbus, OH. I am about to start my 3rd year at the Ohio State University studying math, with a minor in Hebrew. I have an older sister who lives in New York and at home I have two dogs. Please comment the posts if you have any questions or comments. Thanks!