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$\lim \limits_{x \to e}\space\space(1-\log x)\log (x-e)$

How can is it possible to eliminate indeterminate without using L'Hospital's Rule ? I tried to manipulate this formula but still the same problem.

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  • $\begingroup$ We need $x>e,$ Set $y=x/e$ $$\lim_{y\to1^+} \{1-\ln(ey)\}\ln\{e(y-1)\}=-\lim_{y\to1^+}\ln y\{1+\ln(y-1)\}$$ $$=-\lim_{y\to1^+}\ln y\ln(y-1)$$ $\endgroup$ – lab bhattacharjee Nov 9 '16 at 16:42
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Let put $$x=te$$ and compute the limit

$$\lim_{t\to1^-}(1-log(te))log(e-te)$$

$$=-\lim_{t\to1^-}log(t)(1+log(1-t))$$

$$=-\lim_{t\to1^-}log(1+(t-1))(1+log(1-t))$$

$$=-\lim_{t\to1^-} \frac{log(1+(t-1))}{t-1}(t-1+(t-1)log(1-t))$$

$=-1(0-0)=0$.

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By mean value theorem, for $x>e$ there exists $\xi_x\in(e,x)$ such that $$1-\log x=\log e-\log x=\frac{1}{\xi_x}(e-x).$$ Note that $$0<\frac{1}{\xi_x}<\frac{1}{e}.$$ Thus as $x$ gets sufficiently close to $e$ from right, it follows that $$0\leq(1-\log x)(\log(x-e))=\frac{1}{\xi_x}[(e-x)\log(x-e)]\leq\frac{1}{e}[(e-x)\log(x-e)].$$ Then use the fact $$\lim_{t\to 0^+}t\log t=0$$ and squeeze law, we have $$\lim_{x\to e^+}(1-\log x)(\log(x-e))=0.$$

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