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Math Help - Using the Squeeze Theorem to find a limit.

  1. #1
    Member ilikedmath's Avatar
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    Exclamation Using the Squeeze Theorem to find a limit.

    We were given the hint to try using the Squeeze Theorem in order to find the limit of the sequence:

    as n approaches ∞.
    The expression can also be written as (taking the nth root of the expression).
    I understand the concept of the squeeze theorem that I need to find functions greater and less than that limit to the same value so I can conclude that limits to that same value as well.
    I don't know how to come up with those functions. It has been 2 years since I last took a calculus class, so I am very rusty with limits.

    So far all I have is that 0 ≤ ≤ 2^n + 3^n.
    So I can say 0 limits to 0, but then how would I evaluate
    the limit of 2^n + 3^n as n approaches ∞? It would just keep getting bigger so I would have that limit as ∞. So I am stuck

    Any help, tips, corrections, and/or suggestions is greatly appreciated.
    Thank you for your time!
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  2. #2
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    Quote Originally Posted by ilikedmath View Post
    We were given the hint to try using the Squeeze Theorem in order to find the limit of the sequence: \left \{(2^n + 3^n)^{\frac{1}{n}} \right \} as n approaches ∞.
    we have: 3 \leq \sqrt[n]{2^n + 3^n} \leq 3 \sqrt[n]{2}. thus: \lim_{n\to\infty} \sqrt[n]{2^n + 3^n} = 3. in general if 0 \leq a_1 \leq a_2 \leq \cdots \leq a_k, then: \lim_{n\to\infty} \sqrt[n]{a_1^n + a_2^n + \cdots + a_k^n} = a_k, because: a_k \leq \sqrt[n]{a_1^n + a_2^n + \cdots + a_k^n} \leq a_k\sqrt[n]{k}. \ \ \Box

    (*) to be tried by more advanced calculus students: the above limit has an interesting "continuous" version, i.e. if f: [a,b] \longrightarrow \mathbb{R} is continuous, then: \lim_{n\to\infty} \left(\int_a^b |f(x)|^n dx \right)^{\frac{1}{n}} = \max_{a \leq x \leq b}|f(x)|.
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