This is my teacher's work. "
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How does the the $a^2\ln(a)/2)$ lose the $1/2$ part when its limit is taken? I.E. the step following $a^2\ln(a)/2$ is limit as a approaches 0 from the right side of $a\ln(a)$. Where did the $1/2$ go?
The derivative of $\ln(u)$ is $du/dx(1/x)$. takin the derivative of $\ln(a)$, where $a$ is a constant, should yield zero. but that is not so in the last steps of taking the limit as $x$ approaches 0.
The fractional $\frac{1}{2}$ isn't really "lost", it just looks like your teacher "stepped out" of the integral temporarily to evaluate the limit $\lim_{a \to 0^+} a^2\ln(a)$. Once your teacher established that $$\lim_{a \to 0^+} a^2\ln(a) = 0$$ you simply use the fact that $$\lim_{a \to 0^+} \left(c \cdot a^2\ln(a) \right) =c \cdot \left( \lim_{a \to 0^+} a^2\ln(a)\right) \\ = c \cdot 0 \\ = 0$$ for any constant number $c$. In your case, $c = \frac{1}{2}$, but again it doesn't matter what coefficient is attached to the limit.