Question

Use polar coordinates to find the limit if it exists

Answer 1

\[\lim_{(x,y) \rightarrow (0,0)}\left( (x^2+y^2)\ln(x^2+y^2) \right)\]

Answer 2

Polars work nice for this since \[r^2 =x^2 + y^2 \] but we can convert to:
\[r^2ln(r^2) \] , notice, r^2 has to approach zero, or just r, so the limit doesn't even depend on theta at all:

\[\lim_{r \to 0} r^2ln(r^2) = 0*-\infty \] (excuse thehand waving). This is indeterminate, so we just use L'hoptials rule and get:
\[\lim_{r \to 0} \frac{ln(r^2)}{\frac{1}{r^2}} \\ \lim_{r \to 0} \frac{\frac{2r}{r^2}}{\frac{-2}{r^3}} \]
Simplifies to
\[\lim_{r \to 0} {\frac{2}{r}} \frac{r^3}{-2} = \lim_{r \to 0} -r^2 = 0 \]

Answer 3

Sorry, my computer went stupid I can't understand the work you typed thru the equation button

Answer 4

It appears to be very laggy, it may have trouble loading the latex, maybe wait to see if it stops, but if not here:

Answer 5

Can you help me with another one?

Answer 6

I can try