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Math Help - Limits at infinity

  1. #1
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    Limits at infinity

    Hey, this question is simple enough, I know the answer. But I need a justification for it.

    As x -> infinity, f(x) = xa^x -> 0. ( if 0 < a < 1)

    I can justify that a^x goes to 0 when x goes to infinity, but that other x term is making it a little more complicated to justify. Do i just say that a^x is getting smaller at a rate faster than x is increasing?
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  2. #2
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    Quote Originally Posted by hashshashin715 View Post
    Hey, this question is simple enough, I know the answer. But I need a justification for it.

    As x -> infinity, f(x) = xa^x -> 0. ( if 0 < a < 1)

    I can justify that a^x goes to 0 when x goes to infinity, but that other x term is making it a little more complicated to justify. Do i just say that a^x is getting smaller at a rate faster than x is increasing?
    0 \times \text{anything} = 0

    so

    0 \times \infty = 0
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  3. #3
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    Are you allowed to use the ratio test?

    \lim_{n \to \infty} \frac{(n+1)a^{(n+1)}}{na^n} =

    \lim_{n \to \infty} \frac{(n+1)}{n} \cdot \frac{a^{(n+1)}}{a^n} =

    \lim_{n \to \infty} \frac{(n+1)}{n} \cdot \lim_{n \to \infty} \frac{a^{(n+1)}}{a^n} =

    1 \cdot a = a

    And since 0 < a < 1,

    \lim_{n \to \infty} na^n = 0
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  4. #4
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    Quote Originally Posted by pickslides View Post

    0 \times \infty = 0
    Not true in general.
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