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Does this limit exist and is finite? (It goes to one from the left

$\lim_{x\to 1^-}{{(\ln(x) \times \ln(ln(x))}})$

I'm not sure about what the $\lim_{x\to 1^-}{{(\ln(ln(x))}})$ returns here. What do I do in this case?

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    Since you tagged the question [tag:calculus], I guess that $\ln$ is only defined for $x > 0$. But if $0 < x < 1$ then $\ln x < 0$, so $\ln (\ln x)$ is not defined. – 2017-02-14
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    you have to specify what you mean by a logarithm of negaive number...are you sure the limit isn't from the right? – 2017-02-14
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    Yes, I'm quite sure. I guess it is a trick quesion – 2017-02-14

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$\ln(\ln(x))$ is only defined for $x>1$, because if $x<1$, then $\ln(x)<0$ which means that $\ln(\ln(x))$ is not defined.

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As noted by others if $x \to 1^{-}$ then the expression $\log\log x$ is not real and hence the limit does not exist as far as we are dealing with real functions of a real variable. On the other hand if $x \to 1^{+}$ then putting $t = \log x$ we see that $t \to 0^{+}$ and then $\log x \cdot \log \log x = t\log t \to 0$.

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    Yet another instance of anonymous downvote. Thanks to the downvoter for this wonderful gesture! – 2017-02-16