Doing some review for a Calc III Midterm and I'm confused by the algebra in this Question:
Calculate the limit or show that it does not exist:
$\lim _{x,y\to > \left(0,0\right)}\left(x^2+y^2\right)\ln\left(x^2+y^2\right)$
I have the solution to this problem, but I'm not entirely sure what's happening step by step. It starts off trivially enough; using the rule that $r^2=x^2+y^2$, we get:
$r^2\ln\left(r^2\right)$
But the solution I have then rewrites this as $2r^2\ln\left(r\right)$.
Is this correct? I don't understand how the r in the ln function went down a degree - and how the two was added as a coefficient.
I have a follow up question for the rest of the solution: eventually the solution hinges on using L'hopital, but I don't get how this formula can be rewritten as a fraction (unless you use negative exponents, I guess?).
Finally, I'm not sure how this would prove a limit exists (is this a special property of functions that can be re-expressed with polar coordinates?
Yes, it is correct. Note that: $$\ln(a^n)=n\ln(a)$$ Where $a\in \mathbb{R}^+$ and $n \in \mathbb{R}$.
Yes, you should use L'Hopital's rule with negative exponents. Note that: $$\lim_{r\to 0} 2r^2\ln(r)=2\cdot \lim_{r \to 0} \frac{\ln{r}}{\frac{1}{r^2}}$$ Using L'Hopital's rule gives: $$2\cdot \lim_{r \to 0} \frac{\frac{1}{r}}{-\frac{2}{r^3}}=2\cdot \lim_{r \to 0} -\frac{r^2}{2}=\cdots$$