Question

find the limit, if it exists, or type dne if it does not exist.
a. $lim_{(x,y)\to(0,0)}\frac{(x + 24y)^2}{x^{2}+576y^{2}}$
b. $lim_{(x,y)\to(0,0)}\frac{9x^{3}+8y^{3}}{x^{2}+y^{2}}$
(hint for part b: let $x = rcos\theta$, $y = rsin\theta$, and note that $x^{2}+y^{2}=r^{2}$).

Explanation:

Step1: Convert to polar coordinates for part A

Let \(x = r\cos\theta\) and \(y=r\sin\theta\). Then \(x^{2}+y^{2}=r^{2}\) and \((x + 24y)^{2}=(r\cos\theta+24r\sin\theta)^{2}=r^{2}(\cos\theta + 24\sin\theta)^{2}\). The limit \(\lim_{(x,y)\to(0,0)}\frac{(x + 24y)^{2}}{x^{2}+576y^{2}}\) becomes \(\lim_{r\to0}\frac{r^{2}(\cos\theta + 24\sin\theta)^{2}}{r^{2}\cos^{2}\theta+576r^{2}\sin^{2}\theta}=\lim_{r\to0}\frac{(\cos\theta + 24\sin\theta)^{2}}{\cos^{2}\theta+576\sin^{2}\theta}\). Since this limit does not depend on \(r\), we can find its value. Expand the numerator: \((\cos\theta + 24\sin\theta)^{2}=\cos^{2}\theta + 48\sin\theta\cos\theta+576\sin^{2}\theta\). So the limit is \(\frac{\cos^{2}\theta + 48\sin\theta\cos\theta+576\sin^{2}\theta}{\cos^{2}\theta+576\sin^{2}\theta}\). If \(\theta = 0\), the limit is \(\frac{1 + 0+0}{1+0}=1\). If \(\theta=\frac{\pi}{2}\), the limit is \(\frac{0 + 0+576}{0 + 576}=1\). So \(\lim_{(x,y)\to(0,0)}\frac{(x + 24y)^{2}}{x^{2}+576y^{2}} = 1\).

Step2: Convert to polar coordinates for part B

Let \(x = r\cos\theta\) and \(y = r\sin\theta\). Then \(x^{2}+y^{2}=r^{2}\), \(9x^{3}+8y^{3}=9r^{3}\cos^{3}\theta+8r^{3}\sin^{3}\theta\). The limit \(\lim_{(x,y)\to(0,0)}\frac{9x^{3}+8y^{3}}{x^{2}+y^{2}}\) becomes \(\lim_{r\to0}\frac{9r^{3}\cos^{3}\theta+8r^{3}\sin^{3}\theta}{r^{2}}=\lim_{r\to0}r(9\cos^{3}\theta + 8\sin^{3}\theta)\). As \(r\to0\), regardless of the value of \(\theta\), \(r(9\cos^{3}\theta + 8\sin^{3}\theta)\to0\). So \(\lim_{(x,y)\to(0,0)}\frac{9x^{3}+8y^{3}}{x^{2}+y^{2}}=0\).

Answer:

A. 1
B. 0