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I tried:

$$\lim_{x \rightarrow e} \frac{e-x}{\ln x -1} = \frac{e-x}{\ln(x)-\ln(e)} = \frac{e-x}{\ln(\frac{x}{e})} = ???$$

What do I do next? I think I could use $\lim \frac{\log_a x}{x} = 0, a>1$, but I'm not sure how.

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    You can split the numerator and denominator, but I doubt that'll help out. – 2017-02-24
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    If you aren't allowed to use L'Hopital's, are you allowed to use the 'textbook' definition of the derivative itself? – 2017-02-24
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    @StevenStadnicki It depends I think. How would you use it in this case? – 2017-02-24
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    @MarkRead 'Flip over' your middle term, negate the denominator, and it should look very familiar... – 2017-02-24

4 Answers 4

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$$\frac{e-x}{\ln(x)-1}=-\left(\frac{\ln(x)-\ln(e)}{x-e}\right)^{-1} \rightarrow_{x \to e} -(\ln'(e))^{-1}=-e$$ by the very definition of the derivative.

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We'll look at the middle term in particular: let $L=\lim_{x\to e}\dfrac{e-x}{\ln(x)-\ln(e)}$. Then $\dfrac1L = \lim_{x\to e}\dfrac{\ln(x)-\ln(e)}{e-x}$ and so $-\dfrac1L=\lim_{x\to e}\dfrac{\ln(x)-\ln(e)}{x-e}$. This should be a very familiar-looking form if you know the standard textbook definition of the derivative...

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To avoid the derivative (since this is effectively L'Hospital's rule), notice that

$$\lim_{x \rightarrow e} \frac{e-x}{\ln x -1} = -e\lim_{x \rightarrow e} \frac{\frac{x}{e} - 1}{\ln \frac{x}{e} } = -e \lim_{y \to 1} \frac{y-1}{\ln y}.$$

Using the well-known inequality $(y-1)/y \leqslant \ln y \leqslant y - 1$ it follows that

$$1 \leqslant \frac{y-1}{\ln y} \leqslant y,$$

and applying the squeeze theorem

$$\lim_{x \rightarrow e} \frac{e-x}{\ln x -1} = -e \lim_{y \to 1} \frac{y-1}{\ln y} = -e \cdot 1 = -e.$$

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    mmm... I don't think using the definition of the derivative is effectively L'Hopital - not at all. – 2017-03-12
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$$\lim_{x\to e} \frac{e-x}{\ln x-1} = -\lim_{x\to e}\frac{1}{\frac{\ln x-1}{x-e}} = -\lim_{x\to e}\frac{1}{\ln(\frac{x}{e})^{\frac{1}{x-e}}} = -\lim_{x\to e}\frac{1}{\ln{\bigg({\big(1+\frac{x-e}{e}\big)}^{\frac{e}{x-e}}\bigg)^{\frac{x-e}{e}\cdot\frac{1}{x-e}}}} = -\frac{1}{\ln e^{\frac{1}{e}}} = -e$$ With the well-known $$\lim_{f(x)\to 0}(1+f(x))^\frac{1}{f(x)} = e$$