$\lim_{(x,y)\to (0,0)} \frac{\sin(x^2+y^2)}{x^2+y^2}$

Question

I must admit that I've forgotten how to do multivariable limits. Nevertheless I need to know whether the following exists: $$\lim_{(x,y)\to (0,0)} \frac{\sin(x^2+y^2)}{x^2+y^2}$$

Would it be as simple as defining a function $(x^2+y^2)\mapsto z$. Then $$\lim_{(x,y)\to (0,0)} \frac{\sin(x^2+y^2)}{x^2+y^2} = \lim_{z\to 0^+} \frac{\sin(z)}{z} = 1?$$

André Nicolas · Accepted Answer · 2016-05-22T22:26:20.760

2

Hint: Change to polar coordinates and use a well-known limit. One does not even need to go to polar coordinates, but that is a useful move in general when the denominator is $x^2+y^2$. (Your solution is correct.)

edited May 22 '16 at 22:26

answered May 22 '16 at 22:24

André Nicolas

507,029

So $\lim_{(x,y)\to (0,0)} \frac{\sin(x^2+y^2)}{x^2+y^2} = \lim_{r\to 0} \frac{\sin(r^2)}{r^2}$? – user341126 May 22 '16 at 22:25
Yes, and that limit is $1$, as you know. – André Nicolas May 22 '16 at 22:29
@user341126: You are welcome. You may have forgotten a bit about $2$ variable limits, but as you saw, this one is essentially a one variable problem, and you have not forgotten about those. – André Nicolas May 22 '16 at 22:34

score 1 · Answer 2 · answered May 22 '16 at 22:25

1

Hint. You may use polar coodinates, $x=r \cos \theta$, $y=r \sin \theta$, then you initial function writes $$ \frac{\sin (r^2)}{r^2} $$ and consider $r \to 0$.

answered May 22 '16 at 22:25

Olivier Oloa

120,989

$\lim_{(x,y)\to (0,0)} \frac{\sin(x^2+y^2)}{x^2+y^2}$

2 Answers2