Find $\lim\limits_{t \to 0} \int\limits_{2^t}^{3^t} \frac{x}{\ln x} dx$. I have the following limit to find:
$$\lim_{t \to 0} \int _{2^t} ^ {3^t} \dfrac{x}{\ln x} dx$$
This wouldn't be a problem if instead of the given fraction, we would have $\dfrac{\ln x}{x}$ since then I could a substitution like $t = \ln x$, but it looks like that doesn't work here.
Most of the limit-integrals that I worked with were solved with the Squeeze Theorem, but I don't see any boundaries that I could use here.
 A: Let's put $x=u^t$ so that the integral reduces to $$\int_{2}^{3}\frac{u^{2t-1}}{\log u} \, du$$ As $t\to 0$ we can switch limits and integral (because integrand is continuous as a function of two variables) and get the result as $$\int_{2}^{3}\frac{du}{u\log u} =\log\log 3 - \log\log 2$$
A: $$\int_{e^{t\log 2}}^{e^{t\log 3}}\frac{x}{\log x}\,dx \stackrel{x\mapsto e^z}{=} \int_{t\log 2}^{t\log 3}\frac{e^{2z}}{z}\,dz\stackrel{z\mapsto t u}{=}\int_{\log 2}^{\log 3}\frac{e^{2tu}}{u}\,du$$
When $t\mapsto 0$, by the dominated convergence theorem the RHS goes to
$$ \int_{\log 2}^{\log 3}\frac{du}{u}=\color{red}{\log\log 3-\log \log 2}.$$
A: For small $t$,
we have
$a^t 
=e^{t\ln(a)}
\approx 1+t\ln(a)
$
so
$\begin{array}\\
 \int _{2^t} ^ {3^t} \dfrac{x}{\ln x} dx
&\approx  \int _{1+t\ln(2)} ^ {1+t\ln(3)} \dfrac{x}{\ln x} dx\\
&=  \int _{t\ln(2)} ^ {t\ln(3)} \dfrac{x+1}{\ln (1+x)} dx\\
&\approx  \int _{t\ln(2)} ^ {t\ln(3)} \dfrac{x+1}{x} dx\\
&=  \int _{t\ln(2)} ^ {t\ln(3)} (1+\dfrac1{x})dx\\
&= (1+\ln(x))_{t\ln(2)} ^ {t\ln(3)}\\
&= (t\ln(3)-t\ln(2))+(\ln(t\ln(3))-\ln(t\ln(2)))\\
&= (t\ln(3)-t\ln(2))+(\ln(t)+\ln(\ln(3))-\ln(t)-\ln(\ln(2)))\\
&= (t\ln(3)-t\ln(2))+(\ln(\ln(3))-\ln(\ln(2)))\\
&= t\ln(3/2)+(\ln(\frac{\ln(3)}{\ln(2)})\\
&\to\ln(\frac{\ln(3)}{\ln(2)})\\
\end{array}
$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\lim_{t \to 0}\int_{2^{\large t}}^ {3^{\large t}}{x \over \ln\pars{x}}\,dx
& =\lim_{t \to 0}\int_{2^{\large t}}^ {3^{\large t}}{dx \over x - 1} =
\lim_{t \to 0}\ln\pars{3^{t} - 1 \over 2^{t} - 1} =
\ln\pars{\lim_{t \to 0}{3^{t} - 1 \over 2^{t} - 1}}
\\[3mm] & =
\ln\pars{\lim_{t \to 0}{3^{t}\ln\pars{3} \over 2^{t}\ln\pars{2}}} =
\bbox[15px,#ffc,border:1px groove navy]
{\ln\pars{\ln\pars{3} \over \ln\pars{2}}}\ \approx\
0.4606
\end{align}
