$f(x)= \begin{cases} 1, & \text{if $n$ is rational number} \\[2ex] -1, & \text{if $n$ is irrational number} \end{cases}$

$\lim\limits_{x \to 1} \ln(x^{\lvert x\rvert}).\exp(f(x))$

I know that f(x) is bounded, in first look it seems that the limit is zero, but I suspect of the limit existence because of the natural logarithm domain.

  • 1
    $\begingroup$ ...and the question is...? Where's the self work, effort, insights....? That $\;|x|\;$ power is on the logarithm's argument or on the logarithm itself? $\endgroup$ – Timbuc Apr 3 '15 at 8:51
  • $\begingroup$ I amended it :) $\endgroup$ – Alireza Ghaffari Apr 3 '15 at 8:56
  • $\begingroup$ As you know $\lim\limits_{x\to1} |x| \log x = 0$ and like $\lim\limits_{x\to1}e\cdot|x| \log x = 0$ and $\lim\limits_{x\to1}e^{-1}\cdot |x| \log x = 0$ as well. You might use the definition of limit by sequences to show, that the limit of your function exists. $\endgroup$ – Leonhardt von M Apr 3 '15 at 9:02

Because $x^{|x|}$ is continuous in $1$, $$\lim_{x\to1}x^{|x|}=1$$ and hence $$\lim_{x\to1}e^{f(x)}\cdot\ln(x^{|x|})$$ exists and equals $0$ because $e^{f(x)}$ is bounded and $\lim_{x\to1}\ln(x^{|x|})=\ln\left(\lim_{x\to1}x^{|x|}\right)=0$.

  • $\begingroup$ Does $u^v$ if u is negative, defiens for $v \notin z+?$ $\endgroup$ – Alireza Ghaffari Apr 3 '15 at 9:35
  • 1
    $\begingroup$ There are some definitions using the complex logarithm, but that isn't relevant here because we only need $u$'s in a neighbourhood of $1$. $\endgroup$ – punctured dusk Apr 3 '15 at 9:37

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