Compute the following limit:
$\lim_{(x,y)\rightarrow (0,0)}\frac{x^2-y^2}{\sqrt{x^2+y^2}}$
Asked
Active
Viewed 111 times
4
-
are you familiar with polar conversion? – imranfat Mar 17 '14 at 20:29
-
Yes. Please let me know... – Bittu Mar 17 '14 at 20:30
-
Ok, thanks I got it. thanks again. – Bittu Mar 17 '14 at 20:31
-
1If you substitute $x=rcost$ and $y=rsint$ into the problem, what happens? – imranfat Mar 17 '14 at 20:32
2 Answers
2
$$\begin{cases}x=r\cos t\\y=r\sin t \end{cases}\implies \frac{x^2-y^2}{\sqrt{x^2+y^2}}=r\cos 2t\xrightarrow[r\to 0]{}0$$
DonAntonio
- 211,718
- 17
- 136
- 287
0
$0\leq|\frac{x^2-y^2}{\sqrt{x^2+y^2}}|\leq\frac{x^2+y^2}{\sqrt{x^2+y^2}}\leq |x|+|y|\rightarrow 0 $ when $(x,y)\rightarrow (0,0)$. Thus, the limit is zero.
juank2016
- 69