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Limit at Infinity

  1. May 14, 2015 #1
    1. The problem statement, all variables and given/known data
    $$\lim_{x\to\infty} \dfrac{(-1)^n\sqrt{n+1}}{n}$$

    2. Relevant equations
    3. The attempt at a solution

    This is what I managed to do but I just wanted to verify that this is the correct way of solving it, I'm mainly concerned about the fact that I took the absolute value with the log function, is that a valid operation?
    $$y=\lim_{x\to\infty} \dfrac{(-1)^n\sqrt{n+1}}{n} $$
    $$ \ln y=\lim_{x\to\infty} \ln|\dfrac{(-1)^n\sqrt{n+1}}{n}| $$
    $$ \ln y=\lim_{x\to\infty} \ln|\dfrac{(-1)^n\sqrt{n+1}}{n}| $$
    $$\ln y=\lim_{x\to\infty} \ln|(-1)^n|+\lim_{x\to\infty} \ln|\dfrac{\sqrt{n+1}}{n}|$$
    $$\ln y=\lim_{x\to\infty} \dfrac{\ln|1|}{n^{-1}}+\lim_{x\to\infty} \ln|{\sqrt{1/n+1/n^2}}|$$
    $$\ln y=\lim_{x\to\infty} \dfrac{0}{n^{-2}}+ \ln|0|$$
    $$\ln y=-\infty$$
  2. jcsd
  3. May 14, 2015 #2


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    Homework Helper

    There's a whole lot else wrong in there too. What's with the limit ##x \to \infty## when there is no ##x## in the expression? Assume you meant ##n \to \infty##. And there's not need to take a log to begin with. You seem ok with ##lim_{n \to\infty} \dfrac{\sqrt{n+1}}{n}=0##. The ##(-1)^n## doesn't change that much. Just use a squeeze argument.
  4. May 14, 2015 #3
    Could you explain whats wrong other than the fact I accidentally used x->infinity so I don't make that mistake again?
  5. May 14, 2015 #4


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    Not as much as I thought at first, but the limit of ##x_n## is not necessarily the same as ##|x_n|##. I would just skip the log and absolute value in the argument altogether.
  6. May 15, 2015 #5


    Staff: Mentor

    Ignoring the (-1)n factor for the moment, you have
    ##\frac{\sqrt{n+1}}{n} = \frac{\sqrt{n}\sqrt{1 + 1/n}}{\sqrt{n}\sqrt{n}} = \frac {\sqrt{1 + 1/n}}{\sqrt{n}}##
    Can you take the limit now?

    For the original problem, use the squeeze theorem that Dick suggests. I agree that logs and absolute values are not needed.
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