Problem with a chart

1. Aug 15, 2006

Johnnycab

Hi everyone im COmpletly oblivious when it comes to calculus and im going to try and learn it myself. i took a calculus course in college and got a C, i remember some things but i pretty much cheated. That was a couple of years ago, and well now i want to learn it . Anyways here is my problem

This is an example, not my own work

lim (x^2 - 4) / (5x - 10)
x->2

when i sub 2 into x i get 0/0, so im asked to make a chart im given

x **********(x^2 - 4) / (5x - 10) right side limit
3 ****1.0
2.5 ****** 0.9
2.1 ****** 0.82
2.01 ******* 0.802
2.001 ********* 0.8002

x ********(x^2 - 4) / (5x - 10) left side limit
1 ******* 0.6
1.5 ******* 0.7
1.9 ****** 0.78
1.99 ***** 0.798
1.999 ****** 0.7998

what i dont understand is how the book got these numbers, like on the right side 3 appears to equal 1 and on the left side 1 appears to equal (.6). Im saying appears to equal because i didnt see a equal sign. what i think i understand is that on the left side its showing x getting closer to a. But i dont know :surprised

I hope this is enough information on my part to count, i would really appreciate someone to help me understand

PS- the astriks are there to make space between the numbers

Last edited: Aug 15, 2006
2. Aug 15, 2006

d_leet

They are showing you x values around 2, the value that you really care about, and then showing you what the function evaluates to for those values of x.

3. Aug 15, 2006

Johnnycab

that sounds right i guess, the answer ends up being 0.8, and not 0/0. How the auther of the example got that i wont know.

Could i just factor the equation instead?

thank you for your feedback btw :)

4. Aug 15, 2006

Integral

Staff Emeritus
All that the book has done is evaluate the given function at the specified points. Try it. It is not clear to me why you have trouble with f(1)=.6 and f(3)=1, separate points, separate results. The best way to see this is to plug in the numbers for yourself.

5. Aug 15, 2006

D_Dean

0/0 is undefined in our mathematics. In the event of an undefined solution you need use a different technique. You can either try to factor out stuff or in this case you have to use what is called l'hopital's rule (I am sure your text will go into that in the future). What the author was doing was showing you that as your input values (x) gets closer and closer to 2, the solution gets closer and closer to .8, which will eventually lead to the conclusion that the solution (the limit) is in fact .8.

With limits your not asking what is the value of the equation when you plug in this input (2, in our example), but rather what is it approaching when you approach the input (2).

This can be seen easily by asking the following:

lim (x+1)(x-3)/(x-3)
x->3

If you were to graph this it would look like (x+1) (because the (x-3)/(x-3) terms would factor out) but since there is the (x-3)/(x-3) the solution is undefined (0/0) at x=3 and therefore there is a hole in the line at that value (there is an undefined solution at x=3). But you can see that the equation (x+1) approaches => (3+1) at x = 3 so you could say the limit is 4 at x=3.

Does that make sense (except for the l'hopital's rule that is a little advanced at this stage.)

6. Aug 15, 2006

d_leet

No, factoring works in this case. 5x-10, and x2-4 share a common factor of x-2. So you don't HAVE to use L'hopital's rule.

7. Aug 16, 2006

Johnnycab

-Cool thank you everyone for your input, i very much appreciate it
-so when im doing limits i can solve equations by
1. substituting for x,
2. factoring
3. using a chart
-i really dont like #3 - charts i hope in the future i can just substitute and factor
-what is L'hopital's rule?

8. Aug 16, 2006

d_leet

If you're working through a calculus book it should be presented later in the book, probably as an application of differentiation, I don't think that I can expplain it very well, however, so I will give you the link to the wikipedia page on this rule, and someone else may also be able to explain it.

http://en.wikipedia.org/wiki/L'HÃ´pital's_rule