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Compute the limit (with variable in exponent)

  1. Nov 12, 2011 #1
    1. The problem statement, all variables and given/known data
    Compute the following sequence (show your work)

    (-2)n+3n
    ÷ (-2)n+1+3n+1


    3. The attempt at a solution
    (-2)n+3n
    ÷ (-2)n(-2)+3n3n

    After this I don't know what to do. I know that if I can prove that something smaller than this for all n's converges to +∞, this will converge to +∞. Also, if I can prove that something bigger that this for all n's converges to 0, this will converge to 0. I'm just not sure how to simplify this equation at all.
     
  2. jcsd
  3. Nov 12, 2011 #2

    gb7nash

    User Avatar
    Homework Helper

    Is the problem:

    [tex]lim_{n \to \infty} \frac{(-2)^{n}+3^{n}}{(-2)^{n+1}+3^{n+1}}[/tex]

    ?

    Look at this again.
     
  4. Nov 12, 2011 #3

    Mark44

    Staff: Mentor

    I'm guess that this is the expression (it's not an equation) whose limit you want to find:
    [tex]\frac{(-2)^n + 3^n}{(-2)^{n+1} + 3^{n+1}}[/tex]

    Multiply the numerator by 1/3n+1 and multiply the denominator by the same thing.
     
  5. Nov 12, 2011 #4
    So I get [itex]\frac{(-2)^{n}(1/3)^{n+1}+(1/3)}{(-2/3)^{n+1}+1}[/itex]

    Since (-2)n → +∞; (1/3)n+1 → 0, then the product of the two will converge to 0 and the top will converge to [itex]\frac{1}{3}[/itex]

    (-2/3)n+1 → 0 so the bottom will converge to 1

    So the whole thing will converge to [itex]\frac{1}{3}[/itex]

    Is this correct?
     
  6. Nov 12, 2011 #5

    Mark44

    Staff: Mentor

    Your reasoning is flawed here. If one factor in a product is growing large without bound (i.e., approaching ∞) and the other factor is approaching 0, there is nothing you can say about the product. The indeterminate form [0* ∞] is similar to other indeterminate forms such as [0/0] and [∞/∞].

    For example, would you say that 4n * (1/3)n also has a limit of zero?

    Instead of writing (-2)n * (1/3)n+1, write these together as a single fraction, and then you can take the limit.
    Yes.
     
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