Problem 1
1. For the function \(f(x,y)\), the point \((a,b)\in\mathbb{R}^2\), and the real number \(L\in\mathbb{R}\), give the epsilon-delta definition for \(\lim_{(x,y)\to(a,b)} f(x,y) = L\). (2 points)
Solution. We say that \(\lim_{(x,y)\to(a,b)} f(x,y) = L\) if for every \(\epsilon>0\), there exists a \(\delta>0\) such that \(||(x,y)-(a,b)|| < \delta\) implies that \(|f(x,y)-L| < \epsilon\).
Problem 2
2. For \(\lim_{(x,y)\to(0,0)} \dfrac{x^3y}{x^6+2y^2}\), show that the limit along any line through the origin exists and equals zero, but if we take the limit along the curve \(y=x^3\), the limit is not zero. What conclusion can you draw from this? (4 points)
Solution. First we take the limit along any line through the origin. Let \(y=ax\). Then, we have the limit
\[ \lim_{x\to 0} \frac{ax^4}{x^6+2a^2x^2} = \lim_{x\to 0} \frac{ax^2}{x^4+2a^2} = 0. \]
On the other hand, if we approach along the curve \(y=x^3\), we get \[ \lim_{x\to 0} \frac{x^3\cdot x^3}{x^6+2(x^3)^2} = \lim_{x\to 0} \frac{x^6}{3x^6} = \lim_{x\to 0} \frac{1}{3} = \frac{1}{3} \ne 0. \]
As the limits as we approach along different curves are not the same, we conclude that the limit itself must not exist.Problem 3
3. Use polar coordinates to evaluate the limit \(\lim_{(x,y)\to(0,0)} \dfrac{x^3+x^5}{x^2+y^2}\). (4 points)
Solution. We make the standard polar substitutions \(x=r\cos(\theta)\), \(y=r\sin(\theta)\), resulting in \(r^2 = x^2+y^2\). After this substitution, we have
\[ \lim_{r\to 0,\, \theta\in\mathbb{R}} \frac{r^3\cos^3(\theta)+r^5\cos^5(\theta)}{r^2} = \lim_{r\to 0,\, \theta\in\mathbb{R}} r(\cos^3(\theta)+r^3\cos^5(\theta)) = 0 \]
due to the squeeze theorem.