\( f(x, y, z)=\ln (x+2 y+3 z) \) Evaluate \( \lim _{(x, y) \rightarrow(1,1)} \frac{x y-y}{x^{2}-1}: \) If \( z=f(x, y)=x^{2}+3 x y-y^{2} \), find \( d z \); If \( x \) changed from 2 to 2.04 and \( y \) from 3 to 2.94 , compare the val of \( \nabla z \) and \( d z \).
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To evaluate the limit, we start by plugging in the values near \((1, 1)\): \[ \lim_{(x, y) \to (1, 1)} \frac{xy - y}{x^2 - 1} = \lim_{(x, y) \to (1, 1)} \frac{y(x - 1)}{(x - 1)(x + 1)} = \lim_{(x, y) \to (1, 1)} \frac{y}{x + 1} = \frac{1}{2}. \] Next, to find \(dz\), we start from \(z = f(x, y) = x^2 + 3xy - y^2\). The total differential \(dz\) is given by: \[ dz = \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y} dy. \] Calculating the partial derivatives: \[ \frac{\partial z}{\partial x} = 2x + 3y, \quad \frac{\partial z}{\partial y} = 3x - 2y. \] So, \[ dz = (2x + 3y) dx + (3x - 2y) dy. \] Now, substituting \(x=2, y=3\): \[ \frac{\partial z}{\partial x} = 4 + 9 = 13, \quad \frac{\partial z}{\partial y} = 6 - 6 = 0. \] Thus, \[ dz = 13 \cdot dx. \] For \(dx = 0.04\) and \(dy = -0.06\): \[ dz = 13 \cdot 0.04 + 0 \cdot (-0.06) = 0.52. \] Now comparing \(\nabla z\) and \(dz\): \(\nabla z = (13, 0)\), hence \[ \nabla z \cdot (dx, dy) = 13 \cdot 0.04 + 0 \cdot (-0.06) = 0.52. \] They provide similar values since \(dz\) approximates changes in \(z\) under small perturbations in \(x\) and \(y\).