A limit via the mean value theorem for integrals

Mean value theorem. Fix $x>0$. On $[x,3x]$, take $f(t)=\dfrac{1}{e^t}$ (continuous) and $g(t)=\dfrac{1}{t}$ (continuous, positive on $[x,3x]$). There exists $c_x\in[x,3x]$ such that
$$\int_x^{3x}\frac{1}{te^t}\,dt=\frac{1}{e^{c_x}}\int_x^{3x}\frac{1}{t}\,dt=\frac{\ln 3}{e^{c_x}}.$$

Limit. Since $c_x\in[x,3x]$ and $x\to0^+$, we have $c_x\to0^+$, hence $e^{c_x}\to1$. Therefore
$$\lim_{x\to0^+}\int_x^{3x}\frac{1}{te^t}\,dt=\frac{\ln 3}{1}=\ln 3.$$