I need help understanding the concept of right and left lim

subzi

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Sep 5, 2007
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Lim 1 / (x^2-4)
a. x-->2+
b. x-->2-
c. x-->-2+
d. x-->-2-

If a substitute 2 for x I get 1/0 = infinity
Depending upon if it approach from the right or left negative or positive infinity

a. Can I pick a point on the number line like 3 for x 1/(3^2-4) and get some positive number infinity
b. 1 for x get some negative number to infinity
c. Pick 1.99 for x get some negative number to infinity
d. Some positive number to infinity

Is there some other way to look at it
 
This is actually a very poor example to get the concept across.
Lets say that \(\displaystyle x>2\) but is very near 2; therefore \(\displaystyle x^2>4\) but very near 4. Is that clear to you?
If so, then \(\displaystyle x^2- 4 >0\) but is very near 0. Right?
But that means \(\displaystyle \frac{1}{{x^2 - 4}} > 0\quad \& \quad \frac{1}{{x^2 - 4}} \approx \infty\).

Can you workout the others.
 
Looking at the graph may help.

See how it's unbounded as you approach -2 or 2 from the left or right?.

That's because at -2 and 2 we have division by 0. That creates a rift in the space-time continuum of the mathematical universe. :)


limitnq0.jpg
 
Can you please elaborate on it more I don’t think I have concept down. I think X >2 because its coming from the positive side so it has to be greater then zero. Thank you.
 
subzi said:
Can you please elaborate on it more
On which concept(s) do you need "elaboration"? (We can't teach lessons, and the concept of limits was probably at least one chapter in your book, so the best we'll be able to do, probably, is provide you with links. But the number of typoes, etc, in the post makes it difficult to determine what you mean.)

Please be specific. Thank you! :D

Eliz.
 
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