Solve $\lim_{x \rightarrow e} \frac{e-x}{\ln x -1}$ without using L'Hopital's rule 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.
 A: $$\frac{e-x}{\ln(x)-1}=-\left(\frac{\ln(x)-\ln(e)}{x-e}\right)^{-1} \rightarrow_{x \to e} -(\ln'(e))^{-1}=-e$$
by the very definition of the derivative.
A: 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...
A: 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
$$\lim_{x \rightarrow e} \frac{e-x}{\ln x -1}  = -e \lim_{y \to 1} \frac{y-1}{\ln y} = -e \cdot 1 = -e.$$
A: $$\lim_{x\to e} \frac{e-x}{\ln x-1} = -\lim_{x\to e}\frac{1}{\frac{\ln x-1}{x-e}} = -\lim_{x\to e}\frac{1}{\ln(\frac{x}{e})^{\frac{1}{x-e}}} = -\lim_{x\to e}\frac{1}{\ln{\bigg({\big(1+\frac{x-e}{e}\big)}^{\frac{e}{x-e}}\bigg)^{\frac{x-e}{e}\cdot\frac{1}{x-e}}}} = -\frac{1}{\ln e^{\frac{1}{e}}} = -e$$
With the well-known $$\lim_{f(x)\to 0}(1+f(x))^\frac{1}{f(x)} = e$$
