Limit of logarithms exponential $$
\lim_{x\to\infty}\biggl(\frac{\ln(x-2)}{\ln(x-1)}\biggr)^{x\ln x}.
$$
 L'Hopital seems like a very hardcore solutions given the situation.Are the any other options?
 A: Let $$f(x) = \biggl(\frac{\ln(x-2)}{\ln(x-1)}\biggr)^{x\ln x}.$$
Then 
$$
\begin{align*}
\ln f(x) &= x\ln x[\ln \ln (x-2) - \ln \ln (x-1)] \\
&= x \ln x\left[\ln \left(1 + \frac{\ln(1 - 2/x)}{\ln x}\right) -  \ln \left(1 + \frac{\ln(1 - 1/x)}{\ln x}\right) \right]\\
&= x \ln x\left[\ln \left(1 - \frac{2}{x \ln x} + o \left(\frac{1}{x\ln x} \right)\right) - \ln \left(1 - \frac{1}{x \ln x} + o \left(\frac{1}{x\ln x} \right)\right)\right] \\
&= x\ln x \left[- \frac{2}{x \ln x} + \frac{1}{x \ln x} + o \left(\frac{1}{x\ln x} \right) \right] \\
&= -1 + o(1).
\end{align*} 
$$
Therefore $\lim_{x \to +\infty} f(x) = e^{-1}$.
A: Not sure this is what you want but here's a solution
this is the natural log of your limit; the logs here are natural logs
$$ L = \lim_{x\to\infty}{\biggl(x\log{x}\log{\biggl({\frac{\log{x-2}}{\log{x-1}}\biggr)}\biggr)}}$$
$$=\lim_{x\to\infty}{\biggl(x\log{x}\log{\biggl({1+\frac{1}{\frac{\log{x-1}}{\log{\left(\frac{x-2}{x-1}\right)}}}\biggr)}\biggr)}}$$
$$=\lim_{x\to\infty}\frac{\left(\frac{\log{x-2}}{\log{x-1}}\right)}
{\log{x-1}}{\frac{\log{x-1}}{\log{\left(\frac{x-2}{x-1}\right)}}
\biggl(\log{\biggl({1+\frac{1}{\frac{\log{x-1}}{\log{\left(\frac{x-2}{x-1}\right)}}}\biggr)}\biggr)}x\log{x}
}$$
$$=\lim_{x\to\infty}\frac{\log{\left(\frac{x-2}{x-1}\right)}}
{\log{x-1}}{x\log{x}}$$
$$=\lim_{x\to\infty}\log{\left(\frac{x-2}{x-1}\right)}x$$
$$=-1$$
A: Let $L$ be the desired limit so that $$\begin{aligned}\log L &= \log\left\{\lim_{x \to \infty}\left(\frac{\log(x - 2)}{\log(x - 1)}\right)^{x\log x}\right\}\\
&= \lim_{x \to \infty}\log\left(\frac{\log(x - 2)}{\log(x - 1)}\right)^{x\log x}\text{ (by continuity of log)}\\
&= \lim_{x \to \infty}x\log x\log\left(\frac{\log(x - 2)}{\log(x - 1)}\right)\\
&= -\lim_{t \to 0^{+}}\frac{\log t}{t}\log\left(\frac{\log(1 - 2t) - \log t}{\log(1 - t) - \log t}\right)\text{ (putting }x = 1/t)\\
&= -\lim_{t \to 0^{+}}\frac{\log t}{t}\log\left(1 - \frac{\log(1 - t) - \log (1 - 2t)}{\log(1 - t) - \log t}\right)\\
&= -\lim_{t \to 0^{+}}\frac{\log t}{t}\dfrac{\log\left(1 - \dfrac{\log(1 - t) - \log (1 - 2t)}{\log(1 - t) - \log t}\right)}{- \dfrac{\log(1 - t) - \log (1 - 2t)}{\log(1 - t) - \log t}}\left(- \dfrac{\log(1 - t) - \log (1 - 2t)}{\log(1 - t) - \log t}\right)\\
&= \lim_{t \to 0^{+}}\frac{\log t}{t}\dfrac{\log(1 - t) - \log (1 - 2t)}{\log(1 - t) - \log t}\\
&= \lim_{t \to 0^{+}}\frac{\log(1 - t) - \log(1 - 2t)}{t}\dfrac{1}{\dfrac{\log(1 - t)}{\log t} - 1}\\
&= \lim_{t \to 0^{+}}\frac{\log(1 - t) - \log(1 - 2t)}{t}\cdot\frac{1}{0 - 1}\\
&= -\lim_{t \to 0^{+}}\dfrac{\log\left(\dfrac{1 - t}{1 - 2t}\right)}{t}\\
&= -\lim_{t \to 0^{+}}\dfrac{\log\left(1 + \dfrac{t}{1 - 2t}\right)}{t}\\
&= -\lim_{t \to 0^{+}}\dfrac{\log\left(1 + \dfrac{t}{1 - 2t}\right)}{\dfrac{t}{1 - 2t}}\cdot\frac{1}{1 - 2t}\\
&= -\lim_{t \to 0^{+}}\frac{1}{1 - 2t} = -1\end{aligned}$$ and hence $L = 1/e$.
