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?

  • $\begingroup$ Write $\ldots = \ln 2 = \frac{x \ln 2}{x}$, then you can use l'Hospital ("$\infty/\infty$ case"). :-) But seriously, what kind of a question is this? Why would anybody want you to do this using l'Hospital? $\endgroup$ – Hans Lundmark Mar 27 at 9:34
  • $\begingroup$ With $f(x)-g(x)$ in the form $\infty-\infty$, one can use $(1-f(x)/g(x))/(1/f(x))$. If this becomes $0/0$ then you can apply l'Hôpital. In this case it's a nightmare. $\endgroup$ – egreg Mar 27 at 10:05

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.

  • $\begingroup$ I already did this. But the problem is using l'hopital rule $\endgroup$ – stackuser Mar 27 at 8:37
  • $\begingroup$ I used l'hopital's rule here after I pass the limit inside the function $\endgroup$ – Tony S.F. Mar 27 at 8:37

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