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Math Help - evaluating limits

  1. #1
    Junior Member winterwyrm's Avatar
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    evaluating limits

    Use the squeeze theorem to show that lim(n->infinity) (5+2n+7(n^2))^(1/n) = 1
    and lim(n->infinity) (ln(n))^(1/n) = 1

    Thanks so much.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by winterwyrm View Post
    Use the squeeze theorem to show that lim(n->infinity) (5+2n+7(n^2))^(1/n) = 1
    and lim(n->infinity) (ln(n))^(1/n) = 1

    Thanks so much.
    I willl do the second one and you do the first. Ok?


    We know that

    \forall{x}\geq{e}\quad{1}\leq\ln(x)

    Also we know that

    \forall{x}\in\mathbb{R}^+\quad{\ln(x)\leq{x}}

    \Rightarrow\forall{x>e}\quad{1\leq\ln(x)\leq{x}}

    Now remaining inequality we see that

    \forall{x}>e\quad1\leq\ln(x)^{\frac{1}{x}}\leq{x^{  \frac{1}{x}}}

    Now we know that if we define D=\left\{x:x\to\infty\right\}

    That and R=\left\{x:x>e\right\}

    We can see that D\subset{R}

    This implies that

    \lim_{x\to\infty}1\leq\lim_{x\to\infty}\ln(x)^{\fr  ac{1}{x}}\leq\lim_{x\to\infty}x^{\frac{1}{x}}


    Now obviously \lim_{x\to\infty}1=1

    and less obviously we can see that

    \lim_{x\to\infty}x^{\frac{1}{x}}\overbrace{=}^{\te  xt{R/R relationsip}}=\lim_{x\to\infty}\frac{x+1}{x}=1

    \Rightarrow\quad1\leq\lim_{x\to\infty}\ln(x)^{\fra  c{1}{x}}\leq{1}

    \Rightarrow\lim_{x\to\infty}\ln(x)^{\frac{1}{x}}=1  ~\text{By the Squeeze Thereom}\quad\blacksquare
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  3. #3
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    Quote Originally Posted by winterwyrm View Post
    Use the squeeze theorem to show that lim(n->infinity) (5+2n+7(n^2))^(1/n) = 1
    1\leq 5+2n+7n^2 \leq 5n^2+2n^2+7n^2 = 14n^2 \implies 1 \leq (5+2n+7n^2)^{1/n} \leq 14^{1/n} \cdot (n^{1/n})^2.
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