$$ f(x, y, z)=\ln (x+2 y+3 z) $$ Evaluate $$ \lim _{(x, y) ightarrow(1,1)} \frac{x y-y}{x^{2}-1}: $$ 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 $$
abla z $$ and $$ d z $$.

Question

$$ f(x, y, z)=\ln (x+2 y+3 z) $$ Evaluate $$ \lim _{(x, y) ightarrow(1,1)} \frac{x y-y}{x^{2}-1}: $$ 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 $$ 
abla z $$ and $$ d z $$.

UpStudy AI · Accepted Answer

Let's break down the problem step by step.

### Step 1: Evaluate the limit

We need to evaluate the limit:

$$
\lim _{(x, y) \rightarrow(1,1)} \frac{x y - y}{x^{2} - 1}
$$

First, we can simplify the expression in the limit. Notice that as $$(x, y) \rightarrow (1, 1)$$, both the numerator and denominator approach 0, which suggests we can apply L'Hôpital's Rule.

Let's rewrite the limit:

$$
\lim _{(x, y) \rightarrow(1,1)} \frac{y(x - 1)}{(x - 1)(x + 1)}
$$

This simplifies to:

$$
\lim _{(x, y) \rightarrow(1,1)} \frac{y}{x + 1}
$$

Now substituting $$x = 1$$ and $$y = 1$$:

$$
\frac{1}{1 + 1} = \frac{1}{2}
$$

### Step 2: Find $$dz$$

Given $$z = f(x, y) = x^{2} + 3xy - y^{2}$$, we need to find the differential $$dz$$.

The differential $$dz$$ is given by:

$$
dz = \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y} dy
$$

First, we calculate the partial derivatives:

1. **Partial derivative with respect to $$x$$**:

$$
\frac{\partial z}{\partial x} = 2x + 3y
$$

2. **Partial derivative with respect to $$y$$**:

$$
\frac{\partial z}{\partial y} = 3x - 2y
$$

Thus, we have:

$$
dz = (2x + 3y) dx + (3x - 2y) dy
$$

### Step 3: Compare $$\nabla z$$ and $$dz$$

Now, we need to evaluate $$\nabla z$$ and $$dz$$ when $$x$$ changes from 2 to 2.04 and $$y$$ changes from 3 to 2.94.

1. **Calculate $$\nabla z$$** at the point $$(2, 3)$$:

$$
\nabla z = \left( \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y} \right) = (2(2) + 3(3), 3(2) - 2(3)) = (4 + 9, 6 - 6) = (13, 0)
$$

2. **Calculate $$dz$$**:

Using $$dx = 2.04 - 2 = 0.04$$ and $$dy = 2.94 - 3 = -0.06$$:

$$
dz = (2(2) + 3(3))(0.04) + (3(2) - 2(3))(-0.06)
$$

Calculating each term:

- First term: $$ (13)(0.04) = 0.52 $$
- Second term: $$ (0)(-0.06) = 0 $$

Thus,

$$
dz = 0.52 + 0 = 0.52
$$

### Summary of Results

1. The limit is $$\frac{1}{2}$$.
2. The differential $$dz$$ is given by $$dz = (2x + 3y) dx + (3x - 2y) dy$$.
3. At the point $$(2, 3)$$, $$\nabla z = (13, 0)$$ and $$dz = 0.52$$.

This concludes the solution. If you have any further questions or need additional calculations, feel free to ask!
The limit is $$\frac{1}{2}$$. The differential $$dz$$ is $$dz = (2x + 3y) dx + (3x - 2y) dy$$. At the point $$(2, 3)$$, $$\nabla z = (13, 0)$$ and $$dz = 0.52$$.

\( f(x, y, z)=\ln (x+2 y+3 z) \) 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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