Question

use polar coordinates to find the limit. hint: let (x = rcos(\theta)) and (y = rsin(\theta)), and note that ((x,y)\to(0,0)) implies (r\to0). (lim_{(x,y)\to(0,0)}sin(sqrt{x^{2}+y^{2}}))

Explanation:

Step1: Substitute polar - coordinates

Let \(x = r\cos\theta\) and \(y = r\sin\theta\). Then \(\sqrt{x^{2}+y^{2}}=\sqrt{r^{2}\cos^{2}\theta + r^{2}\sin^{2}\theta}=r\sqrt{\cos^{2}\theta+\sin^{2}\theta}=r\) (since \(\cos^{2}\theta+\sin^{2}\theta = 1\)). And as \((x,y)\to(0,0)\), \(r\to0\). The limit \(\lim_{(x,y)\to(0,0)}\sin(\sqrt{x^{2}+y^{2}})\) becomes \(\lim_{r\to0}\sin(r)\).

Step2: Evaluate the limit

We know the well - known limit \(\lim_{u\to0}\frac{\sin u}{u}=1\), and \(\lim_{r\to0}\sin(r)\). We can rewrite \(\sin(r)\) as \(r\cdot\frac{\sin(r)}{r}\). Then \(\lim_{r\to0}\sin(r)=\lim_{r\to0}(r\cdot\frac{\sin(r)}{r})\). Using the product rule for limits \(\lim_{r\to0}(r\cdot\frac{\sin(r)}{r})=\lim_{r\to0}r\cdot\lim_{r\to0}\frac{\sin(r)}{r}\). Since \(\lim_{r\to0}r = 0\) and \(\lim_{r\to0}\frac{\sin(r)}{r}=1\), we have \(\lim_{r\to0}\sin(r)=0\).

Answer:

0