Question: Does the following limit exist, if so what does it equal? $$\lim_{(x, y) \to (0,0)} \frac{x^2 y^5}{2x^4 +3y^{10}}$$
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Solution 1: The limit DOES NOT exist.
Let $ x=y^{5/2}$ $$\lim_{y \to 0} \frac{(y^{5/2})^2 y^5}{2(y^{5/2})^4 +3y^{10}} = \lim_{y \to 0} \frac{y^{10}}{2y^{10} +3y^{10}} = \lim_{y \to 0} \frac{1}{2 +3} = \frac{1}{5} $$
Let $ x=0$ $$\lim_{y \to 0} \frac{(0)^2 y^5}{2(0)^4 +3y^{10}} = \lim_{y \to 0} \frac{0}{0 +3y^{10}} = \lim_{y \to 0} 0 = 0 $$
Thus the limit does not exist
Solution 2: The limit DOES exist.
Change to polar coordinates and find limit as $r \to 0$ $$\lim_{r \to 0} \frac{(r^2cos^2{\theta}) (r^5sin^5\theta)}{2(r^4cos^4{\theta}) +3(r^{10}cos^{10}{\theta})} =\lim_{r \to 0}\frac{r^4}{r^4} \times\frac{r^3cos^2{\theta} sin^5\theta}{2cos^4{\theta} +3r^{6}cos^{10}{\theta}} =\frac{0}{2cos^4{\theta} +0}= 0 $$
Thus the limit does exist and is $0$
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Problem Can someone please tell me which solution is correct (if any) and why the other is wrong?
