Is f(x,y) continuous at (0,0) $ f(x,y) = (x+y)^2\cos\left(\frac{1}{\sqrt{x^2+y^2}}\right)$

Question

Show if the following function of two variables has a limit in (0,0). $$ f(x,y) = (x+y)^2\cos\left(\frac{1}{\sqrt{x^2+y^2}}\right);(x,y)\ne(0,0);f(0,0)=0 $$ I tried to find a limit of the function by substituting y = x: $$ f(x,x) = 4x^2\cos(\frac{1}{x\sqrt{2}}) $$ Then easily by squeeze theorem: $$ \lim_{(x,y)\to(0,0)}f(x,x) = 0 $$ On the other hand, after a little fiddling with it, I've come to this: $$ f(x,y) = \frac{\cos\left(\frac{1}{\sqrt{x^2+y^2}}\right)}{(\frac{1}{\sqrt{x^2+y^2}})^2} + \cos\left(\frac{1}{x^2+y^2}\right)2xy $$ EDIT: Here I thought the left side should go to $\infty$, which is obviously wrong. When f(x,y) approaches (0,0), $\frac{1}{f(x,y)}$ approaches infinity. Thus everything is in order. $$ \lim_{(x,y)\to(0,0)} f(x,y) = 0 $$ EDIT: Also, a very useful and stunningly easy solution (from answer below) is using squeeze theorem with cos. $$ -(x+y)^2 \le (x+y)^2\cos(g(x)) \le (x+y)^2 $$ From that it's easily seen that function is continuous.

Answer 1

HINT

Note that

$$\left|(x+y)^2\cos\left(\frac{1}{\sqrt{x^2+y^2}}\right)\right|\le(x+y)^2$$

then refer to squeeze theorem.