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Math Help - Need help to find left/right limit of a/0

  1. #1
    Member Marconis's Avatar
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    Need help to find left/right limit of a/0

    So I have a pretty good understanding of limits with the exception of one thing. When solving algebraically, as you know the first thing to do is plug in using direct substitution. Well, when you get 0/0, you know you need to do more algebra before plugging in. When you get a number a/0, I get confused how to determine the left and right limit.

    For example:

    lim x--> -2 from the right: (x^2-5) / (x+2).

    After solving and then plugging in, you get -1/0. How does this equate to the limit from the right being negative infinity? Could you also tell me how you'd find the left limit? I understand how limits work, but this a/0 case is throwing me off. Without looking at a graph, I am confused how you determine this. My instructor didn't explain it very clearly, and I thought I understood but after looking at it later I realized I was confused. I asked about this on yahoo answers, but I still did not understand this explanation. I don't think I should be having such a hard time with this.

    Thanks for your help!
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  2. #2
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    as x -> -2 from the left , the denominator (x+2) approaches a small negative value ... (-1)/(small negative) = large positive value

    as x -> -2 from the right , the denominator (x+2) approaches a small positive value ... (-1)/(small positive) = large negative value
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  3. #3
    Member Marconis's Avatar
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    What would you do if it were 1/0. How would things change?
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  4. #4
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    Quote Originally Posted by Marconis View Post
    What would you do if it were 1/0. How would things change?
    does the denominator behave as before or is this a new problem?
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  5. #5
    Member Marconis's Avatar
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    I am just asking as if the denominator behaves as before, just to get an idea.

    I got it from this problem though:

    lim x-->0 (1/x^2)= infinity

    The way I solved it is,
    From the left, you get 1/0 where all negative numbers approach 0. Plugging a negative number into x^2 yields a positive result. 1/positive number is infinity in this case.

    From the right, you get 1/0 where all positive numbers approach 0. This keeps it pos/pos, giving you infinity as well.

    Is this the right logic behind it?
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  6. #6
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    yes ... you have to look at the denominator as x -> whatever value from both sides.
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  7. #7
    Member Marconis's Avatar
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    So for my first example, if it were 1/0....

    From the right it'd be infinity, left it'd be negative infinity...right?
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  8. #8
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    No, you cannot simply say "if it were 1/0". That ignores skeeter's question, "what happens to the denominator". Probably the simplest thing to do with a limit of the form a/0 is to calculate the value for x slightly different.

    For example, \frac{1}{x^2} becomes "1/0" at x= 0. If x= 0.0001 then \frac{1}{x^2}= \frac{1}{0.000001}= 1000000 and if x= -0.0001 then \frac{1}{x^2}= \frac{1}{0.000001}= 1000000 so the limit is " +\infty" from both sides.

    lim x--> -2 from the right: (x^2-5) / (x+2).

    x= -1.9999 is close to -2 "on the right". [tex]\frac{x^2- 5}{x+ 2}= \frac{-1.00039999}{0.0001}= 9998.99960001[tex] so \lim_{x\to -2^+} \frac{x^2- 5}{x+ 2} is positive infinity.

    x= -2.0001 is close to -2 "on the left". \frac{x^2- 5}{x+ 2}= \frac{9989.9992}{-.0001}= -99899992 so \lim_{x\to -2^+} \frac{x^2- 5}{x+ 2} is negative infinity.

    (Note that when we say "the limit is" infinity, positive infinity, or negative infinity, those are all ways of saying there is NO limit for a particular reason.)
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  9. #9
    Member Marconis's Avatar
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    Wow, now that makes a lot of sense! Thank you so much for that. I really get it, now.
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  10. #10
    Member Marconis's Avatar
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    10charnvm
    Last edited by Marconis; September 24th 2010 at 07:18 AM.
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