use polar coordinates to find the limit. hint: let (x = rcos(\theta)) and (y = rsin(\theta)), and note that…
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}}))
Answer
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