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Limits & Rationals help please

  1. Apr 3, 2006 #1
    Hi, I'm having trouble with a couple limits because I'm not sure how to answer them exactly...

    a)lim x-> infinity 10^1/x (sorry i don't know how to use LaTex)

    b) lim x-> 3 x^-2 - 3^-2 / x-3
    Now since this one is 0/0, I tried to factor it but I'm not sure HOW to factor it.
  2. jcsd
  3. Apr 3, 2006 #2


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    I take it you mean 10^(1/x). You could do that as 101/x in html or [tex]10^{\frac{1}{x}}[/tex] in Latex (click on that to see the code. What does 1/x go to as x-> infinity? What is 10 to that power?

    If you don't use Latex, PLEASE use parentheses. I assume you mean
    (x^(-2)- 3^(-2))/(x- 3). Did you consider multiplying numerator and denominator by x^2? That gives you (1- x^2/9)/(x^2(x-3)). If you REALLY don't like fractions multiply numerator and denominator by 9= 3^2. That gives (9- x^2)/(9x^2(x-3)). Can you factor that?
  4. Apr 3, 2006 #3
    1) As x-> infinity, 1/x = 0, therefore 10^0 = 1

    2) How come you multiplied x^2? (Our teacher was teaching us that we factor if, when the limit is put in and you get 0/0, that you should factor)
  5. Apr 3, 2006 #4


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    Because it gives you something you CAN factor! You seemed to be having difficulty with the negative powers, I got rid of them.

    Of course, you know that a^2- b^2= (a-b)(a+b) so you certainly could do x^(-2)- 3^(-2)= (a^(-1))^2- (3^(-1))^2= (a^(-1)- 3^(-1))(a^(-1)+ 3^(-1)) That would also work.
  6. Apr 3, 2006 #5
    Is the final answer -2/27?
  7. Apr 4, 2006 #6


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    That looks good to me. :)
    You can always check it by plugging some value of x in the expression, and if it evaluate to some value closed to your result, then it's correct.
    Say, let's choose x = 2.99 (closed to 3). Plug it in and you will have: -0.07444609..., which is ver closed to -2 / 27. And you are correct.
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