Find the limit $$\lim\limits_{x\to\infty}\int^{2x}_{x}\frac{1}{t}dt$$
My trial
\begin{align}\lim\limits_{x\to\infty}\int^{2x}_{x}\frac{1}{t}dt &= \lim\limits_{x\to\infty}\large\left[\ln t \large\right]^{2x}_{x}\\&= \lim\limits_{x\to\infty}\large\left[\ln 2x -\ln x \large\right]\end{align}
This yields the indeterminate form $\infty-\infty.$ I'm thinking of applying L'Hopital's rule but no headway. Any hints, please?
Best Answer
$\displaystyle \lim_{x \to \infty} \int_x^{2x} \dfrac{dt}{t} = \lim_{x \to \infty} [\ln(2x) - \ln x] = \lim_{x \to \infty}[\ln x + \ln 2 - \ln x] = \lim_{x \to \infty} [\ln 2] = \ln 2. \tag 1$