1. **Evaluate the limit** $$\lim_{(x,y) \to (0,0)} \frac{xy}{x^2 + y^2}$$ - We check the limit along different paths to see if it exists. - Along $y=0$, $$\lim_{x \to 0} \frac{x \cdot 0}{x^2 + 0} = 0$$ - Along $x=0$, $$\lim_{y \to 0} \frac{0 \cdot y}{0 + y^2} = 0$$ - Along $y = mx$, substitute: $$\lim_{x \to 0} \frac{x(mx)}{x^2 + (mx)^2} = \lim_{x \to 0} \frac{m x^2}{x^2 + m^2 x^2} = \lim_{x \to 0} \frac{m}{1 + m^2} = \frac{m}{1 + m^2}$$ - Since this depends on $m$, the limit varies with the path. **Conclusion:** The limit does not exist because it depends on the path. 2. **Determine the limit** $$\lim_{(x,y) \to (1,1)} \frac{x(y-1)}{y(x-1)}$$ - Check iterated limits: - $$\lim_{x \to 1} \lim_{y \to 1} \frac{x(y-1)}{y(x-1)} = \lim_{x \to 1} 0 = 0$$ - $$\lim_{y \to 1} \lim_{x \to 1} \frac{x(y-1)}{y(x-1)} = \lim_{y \to 1} 0 = 0$$ - However, the function is undefined at $(1,1)$ and the limit depends on the path. **Conclusion:** The limit does not exist. 3. **Show that** $$\lim_{(x,y) \to (2,1)} (x^2 + 2x - y) = 7$$ - Substitute directly: $$2^2 + 2(2) - 1 = 4 + 4 - 1 = 7$$ **Hence proved.** 4. **Show that** $$\lim_{x \to 0} \lim_{y \to 0} f(x,y) \neq \lim_{y \to 0} \lim_{x \to 0} f(x,y)$$ where $$f(x,y) = \frac{x - y}{2x + y}$$ - Compute LHS: $$\lim_{x \to 0} \lim_{y \to 0} \frac{x - y}{2x + y} = \lim_{x \to 0} \frac{x - 0}{2x + 0} = \lim_{x \to 0} \frac{x}{2x} = \frac{1}{2}$$ - Compute RHS: $$\lim_{y \to 0} \lim_{x \to 0} \frac{x - y}{2x + y} = \lim_{y \to 0} \frac{0 - y}{0 + y} = \lim_{y \to 0} -1 = -1$$ **Since LHS $\neq$ RHS, the limits do not commute.** 5. **Find first order partial derivatives:** (i) $$z = x^3 + y^3 - 3 a x y$$ - $$\frac{\partial z}{\partial x} = 3x^2 - 3 a y$$ - $$\frac{\partial z}{\partial y} = 3y^2 - 3 a x$$ (ii) $$z = x^2 y - x \sin y$$ - $$\frac{\partial z}{\partial x} = 2 x y - \sin y$$ - $$\frac{\partial z}{\partial y} = x^2 - x \cos y$$ 6. **If** $$u = x \log(x y)$$ where $$x^2 + y^2 + 3 x y = 1$$, find $$\frac{du}{dx}$$ - Differentiate constraint implicitly: $$2x + 2y \frac{dy}{dx} + 3 y + 3 x \frac{dy}{dx} = 0$$ - Rearranged: $$\left(2 y + 3 x\right) \frac{dy}{dx} = -2 x - 3 y$$ - So, $$\frac{dy}{dx} = \frac{-2 x - 3 y}{2 y + 3 x}$$ - Differentiate $$u$$: $$\frac{du}{dx} = \log(x y) + x \left( \frac{1}{x} + \frac{1}{y} \frac{dy}{dx} \right) = \log(x y) + 1 + \frac{x}{y} \frac{dy}{dx}$$ - Substitute $$\frac{dy}{dx}$$: $$\frac{du}{dx} = \log(x y) + 1 + \frac{x}{y} \cdot \frac{-2 x - 3 y}{2 y + 3 x}$$ **This is the expression for $$\frac{du}{dx}$$.**