Limit of logarithms exponential

Question

$$ \lim_{x\to\infty}\biggl(\frac{\ln(x-2)}{\ln(x-1)}\biggr)^{x\ln x}. $$ L'Hopital seems like a very hardcore solutions given the situation.Are the any other options?

Answer 1

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

Answer 2

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 $$ L = \lim_{x\to\infty}{\biggl(x\log{x}\log{\biggl({\frac{\log{x-2}}{\log{x-1}}\biggr)}\biggr)}}$$

$$=\lim_{x\to\infty}{\biggl(x\log{x}\log{\biggl({1+\frac{1}{\frac{\log{x-1}}{\log{\left(\frac{x-2}{x-1}\right)}}}\biggr)}\biggr)}}$$

$$=\lim_{x\to\infty}\frac{\left(\frac{\log{x-2}}{\log{x-1}}\right)} {\log{x-1}}{\frac{\log{x-1}}{\log{\left(\frac{x-2}{x-1}\right)}} \biggl(\log{\biggl({1+\frac{1}{\frac{\log{x-1}}{\log{\left(\frac{x-2}{x-1}\right)}}}\biggr)}\biggr)}x\log{x} }$$

$$=\lim_{x\to\infty}\frac{\log{\left(\frac{x-2}{x-1}\right)}} {\log{x-1}}{x\log{x}}$$

$$=\lim_{x\to\infty}\log{\left(\frac{x-2}{x-1}\right)}x$$

$$=-1$$

Answer 3

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 $L = 1/e$.