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Limits Problem

  1. Feb 10, 2009 #1
    1. The problem statement, all variables and given/known data
    [tex]lim_{x\rightarrow\infty}(\frac{3}{5})^{x}[/tex]


    2. The attempt at a solution
    I know the answer is 0 by using the calculator, but how do I solve this algebraically?
     
  2. jcsd
  3. Feb 10, 2009 #2

    Dick

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    Take the log. It's x*log(3/5). As x->infinity what does that approach?
     
  4. Feb 10, 2009 #3
    I don't see how your method works. If you just add the log, you change the function.

    I also have one more.

    [tex]lim_{x\rightarrow\infty}\frac{\sqrt{3x^{2}+x}-\sqrt{x^{2}+2x}}{x}[/tex]

    The answer is [tex]\sqrt{3}-1[/tex] but I get [tex]\frac{\sqrt{3}-1}{0}[/tex]
     
    Last edited: Feb 11, 2009
  5. Feb 11, 2009 #4
    how do u get a zero???? the limit is towards infinity. Try dividing num and denominator with x. Then apply the limits....

    for the first question.... the algebraic way...???? for any number a < 1 limit (x->inf) a^x = 0 always. and 3/5 < 1.
     
  6. Feb 11, 2009 #5
    If I divide the numerator and denominator by x, wouldn't I end where I started?

    But Dick said that you can use logarithms. Can someone show me how to solve my first question using logs? Dick's method was a little confusing to me.
     
  7. Feb 11, 2009 #6

    HallsofIvy

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    The definition of "continuous function" is that lim f(x)= f(lim x). Logarithm is a continuous function so [itex]lim log((3/5)^x)= log(lim (3/5)^x)[/itex]. The lefthand side is x log(3/5). Since 3/5< 1, log(3/5) is negative and x goes to infinity, x log(3/5) goes to negative infinity. In order that log(A)= negative infinity, A must equal 0. [itex]log(lim (3/5)^x))[/itex]= negative infinity so [itex]lim (3/5)^x= 0[/itex]

    Myself, I would consider it simpler, and perfectly valid to note that since 3/5< 1, [itex]3/5)^2< 3/5[/itex], [itex](3/5)^3< (3/5)^2[/itex] etc. so the limit is 0, as praharmitra said.
     
  8. Feb 11, 2009 #7

    Mark44

    Staff: Mentor

    No, not at all. Another way to say this is to factor x from the numerator and denominator. Since lim x/x = 1, as x --> infinity, what you're left with has the same limit.

    Getting a factor of x out of the numerator entails taking a factor of x^2 out of each term in each radical, and this factor of x^2 comes out of the radical as a factor of x.
     
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