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Calculus, left hand, right hand limits.

  1. Mar 25, 2012 #1
    1. The problem statement, all variables and given/known data

    Its just a general query about problems along these lines...

    f(x)=|x^2+3x-18|/(x-3) and a =3, discuss the limiting behaviour of f(x) as x→a^+, as x→a^- and as x→a.



    2. Relevant equations



    3. The attempt at a solution

    So my basic solution to these types of problems are picking a number very close to 3 on either side of the number line, ie, 2.999 and 3.001 and then calculating to see what f9x) approaches given these values of x. The general answer I get is that x is negative on one side and positive on the other therefore as x→a the lim does not exist since lim x→a^- ≠ x→a^+. Although I have a test coming up and I know these types of questions are going to be involved, but I cant use a calculator in tthe test.

    So my question is, do I need to state what f(x) approaces from both sides or is it sufficient to say its negative, then positive... therefore the lim does not exist.

    The only reason I dont say what it approaches is because its hard to calculate ie what |(2.99999)^2 + 3(2.99999) -18|/ (2.99999 -3), is in my head.

    so 1) do I need to state what it approaches... and 2) is there an easy way to calculate the squares and cubes etc of large decimal numbers is ie (2.99999999)^n?

    Any help or thoughts greatly appreciated!
     
  2. jcsd
  3. Mar 25, 2012 #2
    Try breaking up the numerator, as you would typically do for this type of problem.

    The only problem now is that you have absolute values on the top. You can drop the absolute values to form two different cases where as x approaches 3, you keep the numerator positive. Then you can divide out the denominator and calculate the limit by plugging in x=3. Your intuition in right, the limit doesn't exist because the limits are different from the left than the right. By doing this method you can show exactly how they are different.
     
  4. Mar 25, 2012 #3

    SammyS

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    Without the calculator, it's difficult to tell the sign of [itex]x^2+3x-18[/itex] , near x = 3.

    Try factoring [itex]x^2+3x-18[/itex]. You get [itex](x-3)(x+6)\,.[/itex]

    It should be easy to determine the sign of [itex](x-3)(x+6)[/itex] when x is a little greater than 3 and when x is a little bit less than 3.
     
  5. Mar 26, 2012 #4
    Ok thanks I have worked out that you can cancel out terms in the numerator and denominator and then just plug in the approaching values...
    Thanks for the input
     
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