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

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

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    Since $\ln x$ is continuous at $x = 1$, the limit is simply zero. – 2017-02-14
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    Seems like I didn't write the correct limit. Notice the ln(ln(x)) – 2017-02-14
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    I called a rollback. You should [ask a new question](http://meta.math.stackexchange.com/a/25490/290189) then. – 2017-02-14

3 Answers 3

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The function is defined $(0,\infty)$, continuous and derivable... Why would not exist?

$$\ln(1) = 0 \rightarrow \lim_{x \rightarrow 1^{-}}{\ln(x)\cdot \ln(x)} = 0 \cdot 0 = 0 $$

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    It wasn't the limit I intended. The new one is updated – 2017-02-14
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    Please ignore the previous comment. I've [called a rollback](http://meta.math.stackexchange.com/a/25490/290189) since such an edit had rendered an existing answer *wrong*. – 2017-02-14
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The function $\ln^2(x)$ is continuous at 1, therefore the limit exists and

$$0=\ln^2(1)=\lim_{x\rightarrow 1}\ln^2(x)=\lim_{x\rightarrow 1^-}\ln^2(x)$$

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    Seems like I didn't write the correct limit. I changed it – 2017-02-14
  • 0
    Please ignore the previous comment. I've [called a rollback](http://meta.math.stackexchange.com/a/25490/290189) since such an edit had rendered an existing answer wrong. – 2017-02-14
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The function is nowhere defined for $x\le1$, so no.

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    Sorry, I called a rollback. In the edit history of this question, we see that OP changed his question in a significant way that had rendered other existing answers invalid. In this case, I [called a rollback](http://meta.math.stackexchange.com/a/25490/290189). – 2017-02-14