Precal Blog

Last week in class we were given the the nine “laws” of limits. Here are the last two:

8. limit x->0  (sin x)/(x) = 1

9. limit x->0  (1-cos x)/(x) = 0

We were told to accept #8 as true without any proof (we will learn how to prove it next year in Calculus). We were, however, shown how to prove #9, provided that #8 is true.

((1-cos x)/x) * ((1+cos x)(1+cos x))

= ((1-cos² x)/(x(1+cos x)))

= ((sin² x)/(x(1+cos x)))

= ((sin x * sin x)/((x(1+cos x)))

= ((sin x)/x) * ((sin x)/(1+cos x)

= (as limit x->0) 1 * 0

= 0

Then, we were given a worksheet with trigonometric functions, of which we were to find the limits. To solve problems like this, you have to use factoring, as well as all the trig transformations we learned in section 4-3 of the textbook. Using this, you must transform the function so that it becomes one of the expressions from laws #8 and #9, and then find the limit. These are the common trig identities that you will need to know:

tan x = (sin x)/(cos x) = (sec x)/(csc x)

cot x = (cos x)/(sin x) = 1/(tan x) = (csc x)/(sec x)

sec x = 1/(cos x)

csc x = 1/(sin x)

Here’s #3 from the worksheet:

limit x->0  ((sin x)/(2x²-x))

= ((sin x)/x) * 1/(2x-1)

= ((sin x)/x) * 1/(2(0) – 1)

= 1 * -1

= -1

And here’s #5:

limit x->0  ((sec x-1)/(x sec x))

= (sec x -1)/(x) * cos x

= ((cos x sec x)-1)/(x) * cos x

= ((cos x sec x) – cos x)/(x)

= (1-cos x)/(x)

= 0 (because of law #9)

These three things identify a function as having a limit.

x is the x-value you plug in, f(x) is the y-value you get out, a is the x-value that gives you the limit, L is the y-value of the limit itself.

1. If x is close to a but not equal to a, f(x) is close to L.

2. As x gets closer and closer to a but not equal to a, then f(x) gets closer to L.

3. We can make f(x) as close to L by making x close to a but not equal to a.

Number one means that if the x-value you plug into the equation is close in value to a but not equal, then the y-value you get will be close in value to the limit. Number two means that if the x-value you plug in gets closer in value to a but not equal to it, then the y-value you get is closer in value to the limit. Number three means that you can make the y-value you get out of the equation close to the limit by pluging in an x-value close to a, but not equal to it. Overall, the limit of f(x) as x approaches a is L.

Hello everyone!  Today I’m going to go over all the info you need to solve limits algebraicly!  Lets get started!  Please note that I’m doing the best I can with the notation so when you see `(x->3)` what is in parenthasis is supposed to be underneith the word “Lim”.

Lets start with a simple example.

Lim(x->-7),  (2x+5)

Now since the number we are approaching is -7 we want to plug -7 in for x.  This gives us (-14+5).  Now just do the adding and there you go!  The answer is -9!  Now lets move on to something a little more challenging.

Lim(x->2),  `y+2/y^2+5y+6`

This problem is still the same concept, nothing tricky.  You do the same thing, plug in 2 for x because that is the value we are approaching.  We end up working it down to `4/20` which is the same as `1/4`.  Now we will get to the tricky part.

Lim(x->5),  `x-5/x^2-25`

Now look at this carefully, if we plug 5 in for x we will end up with zero in the denominator (Undefined).  This means we need to give the equation a little make over to make so that the denominator is not zero.  To do this you need to factor out the denominator the same you would with a quadratic equation.  After factoring I got `(x-5)/(x+5)(x-5)` for my new formula.  Now you see that there is a (x-5) in both the numberator and the denominator, you can go ahead and cancle those out leaving a 1 in the numberator.  Your final equation should look like `1/(x+5)`.  Now you can plug 5 in for x and you get the answer 1/10.

So always make sure you plug in the value we are approaching in for x.  If this causes a zero in the denominator, look at the equation again and try to factor out the denominator.  In some really tricky problems you will need to factor the numberator too so don’t count that out.  This is important, no matter what you do, do not let the denominator of the equation be zero.  If you work it out, you will be on your way to the answer!

Have a good week everyone!  Summer is almost here!  =]

Hey everyone,

I know that we went over rational functions a while ago, but i was working on my semester portfolio and found that i needed a little help reviewing rational functions.  I found this site online that is really helpful.  It explains rational functions and the other parts that come with them like finding zeros, discontinuities, and asymptotes.  There is also a fun graph where you can manipulate the different parts of a rational function and see what happens graphically as you do so.

Ive attached the link here
Again, its a great review for any missed learning targets or for your semester portfolio!

=]] though i would share it with everyone

This week has been focused on the curriculum for next year, beginning with limits. A limit is a point that a function converges to from BOTH SIDES. We’ve been given three worksheets so far about limits, and Mr. Bieniek asks that we keep them for next year. This is so Ms. Sarnow won’t have to review too much with us in the fall.

If you’re having some trouble understanding limits, here’s a link to an online turorial on limits. It’s very helpful and an almost perfect summary of what we’ve covered so far.

http://archives.math.utk.edu/visual.calculus/1/limits.16/tut1-flash.html

Ok so for the past couple of days in class we have been discussing rational functions. The function y=x+2/x^2-x-6 is a rational function because y equals a ratio of two polynomials. What we have been dealing with lately is how these rational functions look once they are graphed. Once given a rational function, we are told to find three things right away; the x-intercept, y-intercept, and asymptote. An asymptote is a straight line (can be vertical or horizontal) that is approached by a given curve as one of the variables in the equation as the curve approaches infinity.

For example, if we were given g(x)=2x+7/3x-4 , the first thing we would have to do is find the three things that I mentioned above. First find the x-intercept. When solving for the x-intercept, ask yourself, what value for x will get me 0 in the denominator? So set up 2x+7=0, and you will find out that you have an x-intercept of -3.5. Next find the y-intercept. To find this value, just substitute 0 for all of the x’s in the function. So 2(0)+7/3(0)-4 will give you a y-intercept of -7/4. To find the VERTICAL asymptote find the value for x that will give you 0 in the denominator. Set up 3x-4=0, and you will see that there will be an asymptote at 4/3. You might be asking yourself, what do I do once I have all of this information? The answer is graph it! So plot points at (-3.5,0) and (0,-7/4). Then where x=4/3, draw a dashed vertical line. This is the vertical asymptote. This is a value that will never be reached in the rational function, because as x gets closer and closer to 4/3, y will get closer and closer to negative infinity. In order to connect the two points, I would suggest plugging in one more value in for x so you can have another point. Once you have done this and connected the points, you have completed the first half of this problem. You have just graphed the LOCAL BEHAVIOR of the rational function. Now on to step 2!

In this second part, you will be graphing the GLOBAL BEHAVIOR of the rational function g(x)=2x+7/3x-4. By solving for the global behavior, you will determine the oblique (slant) or horizontal asymptote of the rational function. To do so, you will factor the denominator into the numerator using long division. I don’t know how to represent long division on the computer, but hopefully most everyone knows how to long divide. Once you have completed this step, you should have 2/3 as the factor and a remainder of 29/3. Going back to the last unit, we know that we can write these terms as 2/3+((29/3)/(3x-4)). I will be honest, I am not confident about what to do next but I think we graph 2/3+((29/3)/(3x-4)) on the same plane as our g(x) rational function. After doing this, we should get a graph up in the first quadrant, and a horizontal asymptote at 2/3. This graph will never touch 2/3 on the y-axis, and nevr touch 4/3 on the x-axis. Like I said, I get a little lost at this part and correct me if I’m wrong. In order to clear some things up, below are a few links to websites about Horizontal and Oblique Asymptotes of rational functions.

Sorry for no pictures, I’m not good with that, but the websites should have some helpful images.

http://people.richland.edu/james/lecture/m116/polynomials/rational.html

http://www.purplemath.com/modules/asymtote3.htm

http://www.susq-town.org/yohe/rational/rational.htm

http://images.google.com/imgres?imgurl=http://www.mathwords.com/a/a_assets/a127.gif&imgrefurl=http://www.mathwords.com/a/asymptote.htm&usg=__F7oYvO1d4mzRi4rd9v3NYWp_cXg=&h=155&w=243&sz=2&hl=en&start=14&um=1&tbnid=D_K1H-vyPGBsfM:&tbnh=70&tbnw=110&prev=/images%3Fq%3Dhorizontal%2Band%2Boblique%2Basymptotes%26hl%3Den%26rls%3Dcom.microsoft:*%26sa%3DN%26um%3D1