Jameson Lucero

2022-07-06

Evaluating $\underset{x\to 1}{lim}\frac{{\int }_{1}^{x}{\mathrm{ln}}^{2}\left(t\right)dt}{\left(x-1{\right)}^{3}}$

Expert

In terms of the evaluation of the limit: applying L'Hôpital's rule once forms
$\underset{x\to 1}{lim}\frac{{\int }_{1}^{x}{\mathrm{ln}}^{2}\left(t\right)\phantom{\rule{thinmathspace}{0ex}}dt}{\left(x-1{\right)}^{3}}=\underset{x\to 1}{lim}\frac{{\mathrm{ln}}^{2}\left(x\right)}{3\left(x-1{\right)}^{2}},$
where the chain rule implies that the derivative of the denominator is
$\frac{d}{dx}\left[\left(x-1{\right)}^{3}\right]=3\left(x-1{\right)}^{2}.$
From here you can apply L'Hôpital's rule two more times to find
$\underset{x\to 1}{lim}\frac{{\mathrm{ln}}^{2}\left(x\right)}{3\left(x-1{\right)}^{2}}=\underset{x\to 1}{lim}\frac{\mathrm{ln}\left(x\right)}{3x\left(x-1\right)}=\underset{x\to 1}{lim}\frac{1}{x\left(6x-3\right)}=\frac{1}{3}.$
To answer your second question, the result follows from the fundamental theorem of calculus. Let f be a continuous, real-valued function defined on the closed interval [a,b]. Suppose F is defined for all $x\in \left[a,b\right]$ by
$F\left(x\right)={\int }_{a}^{g\left(x\right)}f\left(t\right)\phantom{\rule{thinmathspace}{0ex}}dt.$
Then
${F}^{\prime }\left(x\right)=\frac{d}{dx}{\int }_{a}^{g\left(x\right)}f\left(t\right)\phantom{\rule{thinmathspace}{0ex}}dt=f\left(g\left(x\right)\right){g}^{\prime }\left(x\right).$
Note that we could change the lower integration constant to any other positive real constant $a>0$ (as the fundamental theorem of calculus would still hold) to deduce
${F}^{\prime }\left(x\right)=\frac{d}{dx}{\int }_{a}^{x}{\mathrm{ln}}^{2}\left(t\right)\phantom{\rule{thinmathspace}{0ex}}dt={\mathrm{ln}}^{2}\left(x\right).$
However, if the upper limit of integration was ${x}^{2}$ instead of x, then we would find
${F}^{\prime }\left(x\right)=\frac{d}{dx}{\int }_{a}^{{x}^{2}}{\mathrm{ln}}^{2}\left(t\right)\phantom{\rule{thinmathspace}{0ex}}dt=2x{\mathrm{ln}}^{2}\left({x}^{2}\right).$

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