Limits — Question 4

Question 4 – Solution

Strategy. Bound the numerator using $x^2\le x^2+y^2$ and then compare with a quantity tending to zero.

Step 1: Bound For $(x,y)\ne(0,0)$, $$0\le\frac{x^2y^2}{x^2+y^2}\le\frac{(x^2+y^2)y^2}{x^2+y^2}=y^2.$$ As $(x,y)\to(0,0)$, $y^2\to 0$.

Step 2: Squeeze Both bounding functions tend to $0$, hence $$\boxed{\displaystyle \lim_{(x,y)\to(0,0)}\frac{x^2y^2}{x^2+y^2}=0}.$$

Step 3: Polar check With $x=r\cos\theta$ and $y=r\sin\theta$, $$\frac{x^2y^2}{x^2+y^2}=r^2\cos^2\theta\sin^2\theta.$$ Because $0\le\cos^2\theta\sin^2\theta\le 1/4$, its absolute value is at most $r^2/4\to 0$, uniformly in direction.