Let's solve the **first question (1a)** step by step: ### **Given** \( f(x, y) = \frac{(xy)^2}{(xy)^2 + (x-y)^2} \) We are to compute: - \( L_1 = \lim_{x \to 0} \left(\lim_{y \to 0} f(x, y)\right) \) - \( L_2 = \lim_{y \to 0} \left(\lim_{x \to 0} f(x, y)\right) \) - \( L = \lim_{(x,y) \to (0,0)} f(x, y) \) (if it exists) --- ## **Step 1: Compute \( L_1 \)** \[ L_1 = \lim_{x \to 0} \left(\lim_{y \to 0} \frac{(xy)^2}{(xy)^2 + (x-y)^2}\right) \] **Inner limit**: Fix \(x\), let \(y \to 0\): \[ f(x, y) = \frac{(xy)^2}{(xy)^2 + (x-y)^2} \] As \(y \to 0\): - \((xy)^2 \to 0\) - \((x-y)^2 \to x^2\) So, \[ \lim_{y \to 0} f(x, y) = \frac{0}{0 + x^2} = 0 \] **Outer limit**: \[ L_1 = \lim_{x \to 0} 0 = 0 \] --- ## **Step 2: Compute \( L_2 \)** \[ L_2 = \lim_{y \to 0} \left(\lim_{x \to 0} \frac{(xy)^2}{(xy)^2 + (x-y)^2}\right) \] **Inner limit**: Fix \(y\), let \(x \to 0\): \[ f(x, y) = \frac{(xy)^2}{(xy)^2 + (x-y)^2} \] As \(x \to 0\): - \((xy)^2 \to 0\) - \((x-y)^2 \to y^2\) So, \[ \lim_{x \to 0} f(x, y) = \frac{0}{0 + y^2} = 0 \] **Outer limit**: \[ L_2 = \lim_{y \to 0} 0 = 0 \] --- ## **Step 3: Compute \( L \)** \[ L = \lim_{(x, y) \to (0, 0)} \frac{(xy)^2}{(xy)^2 + (x-y)^2} \] We'll check if the limit exists by approaching along different paths: **Path 1: \(y = 0\)** \[ f(x, 0) = \frac{(0)^2}{(0)^2 + x^2} = 0 \] **Path 2: \(x = 0\)** \[ f(0, y) = \frac{(0)^2}{(0)^2 + y^2} = 0 \] **Path 3: \(y = x\)** \[ f(x, x) = \frac{(x^2 \cdot x^2)}{(x^2 \cdot x^2) + (x-x)^2} = \frac{x^4}{x^4 + 0} = 1 \] Here, the limit along \(y = x\) is 1, but along \(x=0\) or \(y=0\) is 0. **So, the limit does NOT exist.** --- ## **Final Answers** \[ \boxed{ L_1 = 0, \quad L_2 = 0, \quad L \text{ does not exist} } \] --- If you'd like to proceed to part (b) or the next question, let me know!

Let's proceed to **part (b)** of the first question: --- ### **(b) Calculate \(f'(0,0)\) where** \[ f(x, y) = \frac{(xy)^2}{(xy)^2 + (x - y)^2} \] **Note:** The question asks to check whether \(f\) is differentiable at \((0,0)\) and, if so, to find the total derivative \(f'(0,0)\). --- ## **Step 1: Recall the definition of differentiability at \((0,0)\)** \(f\) is differentiable at \((0,0)\) if the limit: \[ \lim_{(h,k) \to (0,0)} \frac{|f(h,k) - f(0,0) - L(h,k)|}{\sqrt{h^2 + k^2}} = 0 \] for some linear map \(L(h,k) = A h + B k\). Since \(f(0,0) = 0\) (from previous limits), the candidate for the derivative is: \[ L(h,k) = a h + b k \] --- ## **Step 2: Compute the partial derivatives at \((0,0)\)** **Partial derivatives:** \[ f_x(0,0) = \lim_{h \to 0} \frac{f(h, 0) - f(0,0)}{h} \] \[ f_y(0,0) = \lim_{k \to 0} \frac{f(0, k) - f(0,0)}{k} \] ### **Compute \(f_x(0,0)\):** \[ f(h, 0) = \frac{(h \cdot 0)^2}{(h \cdot 0)^2 + (h - 0)^2} = \frac{0}{0 + h^2} = 0 \] Thus, \[ f_x(0,0) = \lim_{h \to 0} \frac{0 - 0}{h} = 0 \] ### **Compute \(f_y(0,0)\):** \[ f(0, k) = \frac{(0 \cdot k)^2}{(0 \cdot k)^2 + (0 - k)^2} = \frac{0}{0 + k^2} = 0 \] So, \[ f_y(0,0) = \lim_{k \to 0} \frac{0 - 0}{k} = 0 \] **Therefore, the candidate derivative at \((0,0)\) is:** \[ L(h, k) = 0 \cdot h + 0 \cdot k = 0 \] --- ## **Step 3: Verify the differentiability** We need to check whether: \[ \lim_{(h, k) \to (0,0)} \frac{|f(h,k) - 0|}{\sqrt{h^2 + k^2}} = 0 \] which simplifies to: \[ \lim_{(h, k) \to (0,0)} \frac{|f(h,k)|}{\sqrt{h^2 + k^2}} \] Recall: \[ f(h,k) = \frac{(h k)^2}{(h k)^2 + (h - k)^2} \] --- ## **Step 4: Bound \(f(h, k)\)** From earlier, along different paths, \(f(h,k)\) can be: - \(0\) (along \(h=0\) or \(k=0\)) - \(1\) (along \(h=k\)) We analyze the behavior near \((0,0)\): \[ |f(h,k)| = \frac{(h k)^2}{(h k)^2 + (h - k)^2} \] Note that: \[ (h k)^2 \geq 0 \] \[ (h - k)^2 \geq 0 \] which suggests: \[ |f(h,k)| \leq \frac{(h k)^2}{(h - k)^2} \] when \((h - k)^2 \neq 0\). --- ## **Step 5: Consider the path \(k = m h\)** Let \(k = m h\). Then: \[ |f(h, m h)| = \frac{(h \cdot m h)^2}{(h \cdot m h)^2 + (h - m h)^2} = \frac{(m h^2)^2}{(m h^2)^2 + h^2 (1 - m)^2} \] \[ = \frac{m^2 h^4}{m^2 h^4 + (1 - m)^2 h^2} \] Divide numerator and denominator by \(h^2\): \[ = \frac{m^2 h^2}{m^2 h^2 + (1 - m)^2} \] As \(h \to 0\): \[ |f(h, m h)| \sim \frac{m^2 h^2}{(1 - m)^2} \] which tends to zero. Now, the numerator of our expression: \[ |f(h, m h)| \sim \frac{m^2 h^2}{(1 - m)^2} \] and the denominator: \[ \sqrt{h^2 + (m h)^2} = |h| \sqrt{1 + m^2} \] Thus, \[ \frac{|f(h, m h)|}{\sqrt{h^2 + (m h)^2}} \sim \frac{\frac{m^2 h^2}{(1 - m)^2}}{|h| \sqrt{1 + m^2}} = \frac{m^2 |h|}{(1 - m)^2 \sqrt{1 + m^2}} \] As \(h \to 0\), this tends to **0** for all \(m\). --- ## **Conclusion:** Since along any line \(k = m h\), the ratio tends to zero, and \(f(h,k)\) is bounded appropriately, **the limit is zero**. Therefore, \[ f \text{ is differentiable at } (0,0), \quad \text{with } f'(0,0) = 0 \] --- ## **Final answer:** \[ \boxed{ \text{The function }f \text{ is differentiable at } (0,0), \quad \text{and} \quad f'(0,0) = 0 } \] --- If you'd like to proceed with other parts or questions, let me know!

Let's proceed to **Question (4)**: --- ### **(4) Show that \(F(z,y) = [7]\) (assuming this is a typo, possibly \(F(z,y) = [z, y]\) or similar) and \(X_0 = (1,2)\). Show that \(F\) is differentiable at \(X_0\) with \((dF)_{X_0} = [x]\) by directly verifying** \[ \lim_{(h,k)\to (0,0)} \frac{|F(X_0 + (h,k)) - F(X_0) - (dF)_{X_0}(h,k)|}{\sqrt{h^2 + k^2}} = 0 \] --- ### **Assumptions / Clarification:** Given the notation, it's likely that: - \(F(z, y) = [f(z, y)]\) (possibly a vector-valued function). - At \(X_0 = (1, 2)\), the differential is \((dF)_{X_0} = [x]\), which probably means the differential is the linear map \(L(h,k) = x\), but that seems inconsistent. **Alternatively**, perhaps the problem is: - \(F(z, y) = [z, y]\), the identity mapping from \(\mathbb{R}^2 \to \mathbb{R}^2\). - At \(X_0 = (1, 2)\), the differential is the identity, i.e., the Jacobian matrix is the identity matrix. **Given the information:** - The limit involves verifying the differentiability directly. - The differential at \(X_0 = (1, 2)\) is \([x]\), which might be a typo or shorthand. --- ### **Step 1: Define the function** Suppose \(F(z, y) = (z, y)\), the identity function. At \(X_0 = (1, 2)\): - \(F(1, 2) = (1, 2)\), - The differential \((dF)_{X_0}\) is the identity linear map, i.e., the Jacobian matrix: \[ J_F(1, 2) = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \] which acts on \((h, k)\) as \((h, k)\). --- ### **Step 2: Verify differentiability** We need to verify: \[ \lim_{(h,k)\to (0,0)} \frac{|F(X_0 + (h,k)) - F(X_0) - (dF)_{X_0}(h,k)|}{\sqrt{h^2 + k^2}} = 0 \] Given our assumptions: \[ F(1 + h, 2 + k) = (1 + h, 2 + k) \] \[ F(1, 2) = (1, 2) \] \[ (dF)_{X_0}(h,k) = (h, k) \] So, \[ F(1 + h, 2 + k) - F(1, 2) - (dF)_{X_0}(h,k) = (1 + h - 1, 2 + k - 2) - (h, k) = (h, k) - (h, k) = (0, 0) \] Thus, \[ \frac{|0|}{\sqrt{h^2 + k^2}} = 0 \] which clearly tends to zero as \((h,k) \to (0,0)\). --- ### **### **Conclusion:** The function \(F(z,y) = (z, y)\) (identity map) is differentiable at \((1, 2)\) with derivative: \[ (dF)_{X_0} = \text{identity matrix} \] which maps \((h,k) \to (h,k)\). --- **If the original problem intended a different \(F\), please clarify.** --- ### **Summary:** \[ \boxed{ \text{The function }F(z,y) = (z, y) \text{ is differentiable at } (1,2), \text{ with } (dF)_{X_0} = \text{identity} } \] --- Let me know if you'd like to explore the next question!