Using L'Hospital's rule how can I find $$\lim_{x\to \infty} \int_x^{2x} \frac {1}{t} dt$$
One can easily observe that $\int_x^{2x} \frac {1}{t} dt = \ln(2x)-\ln (x)=\ln2$ so that $\lim_{x\to \infty} \int_x^{2x} \frac {1}{t} dt= \ln 2$. But in this case I am to use L'Hospital's rule. Can I get hints please?
I think one can split $\int_x^{2x} \frac {1}{t} dt$ to $\int_1^{2x} \frac {1}{t} dt-\int_1^x \frac {1}{t}dt$ to obtain the inderminate limit $\infty-\infty$. What transformation can I use to get the indeterminate form $\frac {0}{0}$ so as to apply the L'Hospital's rule?
Best Answer
Maybe they want something like this,
$\lim\limits_{x\to\infty} \int\limits_{x}^{2x} \frac{1}{t}dt = \lim\limits_{x\to\infty} \ln(2x)-\ln(x) = \lim\limits_{x\to\infty}\ln(\frac{2x}{x}) = \ln(\lim\limits_{x\to\infty}\frac{2x}{x}) = \ln(\lim\limits_{x\to\infty}\frac{2}{1}) = \ln(2)$
where we have used that $\ln(x)$ is a continuous function on its domain to pass the limit inside.