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?

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 $\lim_{x \to +\infty} f(x) = e^{-1}$.
