1. **Problem 1:** Interpret the limit \(\lim_{(x,y)\to(0,0)} \frac{x^2 - y^2}{x^3 + y^3}\) by considering limits along coordinate axes. - Along \(y=0\): \[ \lim_{x\to0} \frac{x^2 - 0}{x^3 + 0} = \lim_{x\to0} \frac{x^2}{x^3} = \lim_{x\to0} \frac{1}{x} \to \pm \infty \text{ (does not exist)} \] - Along \(x=0\): \[ \lim_{y\to0} \frac{0 - y^2}{0 + y^3} = \lim_{y\to0} \frac{-y^2}{y^3} = \lim_{y\to0} \frac{-1}{y} \to \pm \infty \text{ (does not exist)} \] Since limits along axes do not exist or differ, the overall limit does not exist. 2. **Problem 2:** Determine if \(\lim_{(x,y)\to(0,0)} \frac{x^4 - 16 y^4}{x^2 + 4 y^2}\) exists and find its value. - Use substitution \(x=0\): \[ \lim_{y\to0} \frac{0 - 16 y^4}{0 + 4 y^2} = \lim_{y\to0} \frac{-16 y^4}{4 y^2} = \lim_{y\to0} -4 y^2 = 0 \] - Use substitution \(y=0\): \[ \lim_{x\to0} \frac{x^4 - 0}{x^2 + 0} = \lim_{x\to0} \frac{x^4}{x^2} = \lim_{x\to0} x^2 = 0 \] - Since both paths give 0, check polar coordinates \(x=r\cos\theta, y=r\sin\theta\): \[ \frac{r^4 \cos^4\theta - 16 r^4 \sin^4\theta}{r^2 \cos^2\theta + 4 r^2 \sin^2\theta} = r^2 \frac{\cos^4\theta - 16 \sin^4\theta}{\cos^2\theta + 4 \sin^2\theta} \] - As \(r \to 0\), the expression tends to 0 regardless of \(\theta\). **Answer:** Limit exists and equals 0. 3. **Problem 3:** Compute \(\lim_{(x,y)\to(0,0)} \frac{e^{-x^2 - y^2} - 1}{x^2 + y^2}\) using polar coordinates. - Substitute \(x=r\cos\theta, y=r\sin\theta\): \[ \lim_{r\to0} \frac{e^{-r^2} - 1}{r^2} \] - Use Taylor expansion for \(e^{-r^2} \approx 1 - r^2 + \frac{r^4}{2} - \cdots\): \[ \frac{(1 - r^2 + \frac{r^4}{2} - \cdots) - 1}{r^2} = \frac{-r^2 + \frac{r^4}{2} - \cdots}{r^2} = -1 + \frac{r^2}{2} - \cdots \] - Taking \(r \to 0\), limit is \(-1\). 4. **Problem 4:** Interpret that mixed second-order partial derivatives of \(f(x,y) = \frac{x^2 - y^2}{x^2 + y^2}\) are equal. - Clairaut's theorem states if \(f\) and its partial derivatives are continuous near a point, then: \[ \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} \] - This means the order of differentiation does not matter for \(f\) where it is defined and smooth. 5. **Problem 5:** Interpret that functions \(u(x,y) = \ln(x^2 + y^2)\) and \(v(x,y) = 2 \tan^{-1}(y/x)\) satisfy the Cauchy–Riemann equations: - The Cauchy–Riemann equations are: \[ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \quad \frac{\partial v}{\partial x} = - \frac{\partial u}{\partial y} \] - Compute partial derivatives: \[ \frac{\partial u}{\partial x} = \frac{2x}{x^2 + y^2}, \quad \frac{\partial u}{\partial y} = \frac{2y}{x^2 + y^2} \] \[ \frac{\partial v}{\partial x} = -\frac{2y}{x^2 + y^2}, \quad \frac{\partial v}{\partial y} = \frac{2x}{x^2 + y^2} \] - These satisfy the Cauchy–Riemann equations, indicating \(u + iv\) is analytic except at \((0,0)\). **Summary:** - Q1: Limit does not exist. - Q2: Limit exists and equals 0. - Q3: Limit equals -1. - Q4: Mixed partial derivatives are equal by Clairaut's theorem. - Q5: \(u,v\) satisfy Cauchy–Riemann equations, so the function is analytic.