Limits — Question 7

Question 7 – Solution

Strategy. The sine factor is bounded even though it oscillates; the radial factor forces the product to zero.

Step 1: Absolute bound Since $|\sin u|\le 1$, $$\left|(x^2+y^2)\sin\!\left(\frac 1{x^2+y^2}\right)\right|\le x^2+y^2.$$

Step 2: Squeeze As $(x,y)\to(0,0)$, the right side tends to $0$. Therefore $$\boxed{\displaystyle \lim_{(x,y)\to(0,0)}(x^2+y^2)\sin\!\left(\frac 1{x^2+y^2}\right)=0}.$$

Step 3: Why oscillation is harmless Along radii for which $1/(x^2+y^2)$ hits alternating sine peaks, the sine values do not settle. Nevertheless, the product’s magnitude never exceeds the shrinking radius squared, so every path is controlled by the same bound.