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How to prove the $\lim\limits_{x\to 1} \frac{1}{\log x}$ does not exist.

I'm not really sure how to proceed with this one. I have two ideas at hand, proof by contradiction or a proof using the Sequential Characterization of Limits (SCL). The SCL is used to illustrate that the limit does not exist by showing two different sequences both converge to the same value but $f(x_n)$ converges to two different values.

Perhaps this problem requires a different approach. If not, I can't seem to find how to apply what I do know to prove it does not exist.

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  • $\begingroup$ $y=\log x$ $$x \to 1 \iff y \to 0 $$ $\endgroup$ – rlartiga Oct 1 '14 at 1:09
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First note that $$ \lim_{x\to 1^-} \frac{1}{\log x} =-\infty $$ And $$ \lim_{x\to 1^+} \frac{1}{\log x} =\infty $$ Therefore $$ \lim_{x\to 1} \frac{1}{\log x} \Rightarrow \mbox{does not exist} $$

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Since $log 1=0$, you can show that if it did have a limit, $L$, for a specific epsilon, say, $1$, you could never have a neighborhood around $1$ such that your values are all within 1 of $L$ (Since you can always get bigger and bigger by taking values closer and closer to 1, then recipricating)

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