Single Variable Calculus, Chapter 7, 7.8, Section 7.8, Problem 84

Suppose that $\displaystyle \nu - c \left( \frac{r}{R} \right)^2 \ln \left( \frac{r}{R} \right)$ where $c$ is constant. Find
$\displaystyle \text{a.) } \lim_{R \to r^+} \nu \qquad \text{b.) } \lim_{r \to 0^+} \nu$



$
\begin{equation}
\begin{aligned}
\text{a.) } \lim_{R \to r^+} - c \left( \frac{r}{R} \right)^2 \ln \left( \frac{r}{R} \right) &= - c \left(\frac{r}{r} \right)^2 \ln \left( \frac{r}{r} \right)\\
\\
&= -c (1)^2 \ln(1)\\
\\
&= -c \cdot 0\\
\\
&= 0
\end{aligned}
\end{equation}
$




$
\begin{equation}
\begin{aligned}
\text{b.) } \lim_{r \to 0^+} - c \left( \frac{r}{R} \right)^2 \ln \left( \frac{r}{R} \right) &= \lim_{r \to 0^+} \left[ \left( \frac{r}{R} \right)^2 \frac{\frac{1}{R}}{\frac{r}{R}} + \ln \left( \frac{r}{R} \right) \cdot 2 \left[ \frac{r}{R} \right] \left[ \frac{1}{R} \right] \right]\\
\\
&= \lim_{r \to 0^+} -c \left[ \left( \frac{r}{R} \right)^2 \frac{1}{R} \left( \frac{R}{r} \right) + 2 \ln \left( \frac{r}{R} \right) \left( \frac{r}{R^2} \right) \right]\\
\\
&= \lim_{r \to 0^+} -c \left[ \frac{r}{R^2} +2 \ln \left( \frac{r}{R} \right) \left( \frac{r}{R^2} \right) \right]\\
\\
&= \lim_{r \to 0^+} -c \left( \frac{r}{R^2} \right) \left[ 1 + 2 \ln \left( \frac{r}{R} \right) \right]
\end{aligned}
\end{equation}
$


If we evaluate the limit, we will still get an indeterminate form, so we must apply the L'Hospital's Rile once more...

$
\begin{equation}
\begin{aligned}
&= \lim_{r \to 0^+} -c \left[ \frac{r}{R^2} \cdot 2 \frac{\left(\frac{1}{R}\right)}{\frac{r}{R}} + \frac{1}{R^2} \cdot 2 \ln \left( \frac{r}{R} \right)\right]\\
\\
&= \lim_{r \to 0^+} -c \left[ \frac{r}{R^2} \cdot 2 \left( \frac{1}{R} \right) \left( \frac{R}{r} \right) + \frac{1}{R^2} \left( 1 + 2 \ln \left(\frac{r}{R}\right)\right) \right]\\
\\
&= \lim_{r \to 0^+} -c \left[ \frac{2}{R^2} + \frac{1}{R^2} + \frac{2}{R^2} \ln \left(\frac{r}{R}\right) \right]\\
\\
&= \lim_{r \to 0^+} \frac{-c}{R^2} \left[ 3 + 2 \ln \left(\frac{r}{R}\right)\right]\\
\\
&= \frac{-c}{R^2} \left[ 3 + 2 \ln \left( \frac{0^+}{12} \right) \right]\\
\\
&= \infty
\end{aligned}
\end{equation}
$