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Thread: Help with limit of sequence

  1. #1
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    Help with limit of sequence

    hello everyone
    I am kind of new in that whole subject of limits of sequences i got those two questions which i didn't manage to solve

    Help with limit of sequence-file.png


    I need to find the limit of each sequence which I found pretty hard since I cannot use the squeeze theorm

    so please I really need help with those two

    thanks
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  2. #2
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    Re: Help with limit of sequence

    For the first one, observe that very "large" n, $\sqrt{n+c}\approx \sqrt n$ for any constant c. So the limit of $S_n$ is 0.

    For the second one, divide numerator and denominator of $b_n$ by $(2/3)^n$ to get
    $$b_n={1\over (3/4)^n+(27/20)^n}$$

    I hope it is now clear that $b_n$ approaches 0.
    Thanks from greg1313
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  3. #3
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    Re: Help with limit of sequence

    Why can you not use the Squeeze Theorem?

    For $S_n$, $6\sqrt{n-3}-6\sqrt{n+7} = 4\sqrt{n-3}-\sqrt{n+7}-5\sqrt{n+7}+2\sqrt{n-3} < S_n < 4\sqrt{n+3}-\sqrt{n-1}-5\sqrt{n-1}+2\sqrt{n+3} = 6\sqrt{n+3}-6\sqrt{n-1}$

    $\displaystyle \begin{align*}\lim_{n \to \infty} \left(6\sqrt{n-3}-6\sqrt{n+7}\right) & = 6\lim_{n \to \infty} \left( \sqrt{n-3} - \sqrt{n+7} \right) \\ & = 6\lim_{n \to \infty} \dfrac{(n-3)-(n+7)}{\sqrt{n-3}+\sqrt{n+7}} \\ & = 6\lim_{n \to \infty} \dfrac{-10}{\sqrt{n-3} + \sqrt{n+7}} = 0\end{align*}$

    Similarly for the upper bound. Apply the Squeeze Theorem.

    Or do you mean the assignment says to not use the Squeeze Theorem?
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  4. #4
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    Re: Help with limit of sequence

    eventually I was requested to used it because it was to complicated to solve it thank you very much for the solution
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