Evaluate the following limit.
ModifyingBelow lim With left parenthesis x comma y comma z right parenthesis right arrow left parenthesis 1 comma 1 comma 4 right parenthesisStartFraction y minus StartRoot yx EndRoot minus StartRoot yz EndRoot plus StartRoot zx EndRoot Over y minus StartRoot yx EndRoot plus StartRoot yz EndRoot minus StartRoot zx EndRoot EndFraction


We are tasked with evaluating the following limit:

\[
\lim_{(x, y, z) \to (1, 1, 4)} \frac{y - \sqrt{yx} - \sqrt{yz} + \sqrt{zx}}{y - \sqrt{yx} + \sqrt{yz} - \sqrt{zx}}.
\]

### Step-by-Step Process:

First, substitute the values \(x = 1\), \(y = 1\), and \(z = 4\) into the expression directly:

\[
y - \sqrt{yx} - \sqrt{yz} + \sqrt{zx} = 1 - \sqrt{1 \cdot 1} - \sqrt{1 \cdot 4} + \sqrt{1 \cdot 4} = 1 - 1 - 2 + 2 = 0,
\]
and
\[
y - \sqrt{yx} + \sqrt{yz} - \sqrt{zx} = 1 - \sqrt{1 \cdot 1} + \sqrt{1 \cdot 4} - \sqrt{1 \cdot 4} = 1 - 1 + 2 - 2 = 0.
\]

Both the numerator and denominator evaluate to zero, indicating that this is an indeterminate form \( \frac{0}{0} \). We will apply **L'Hopital's Rule** to resolve this limit.

### Applying L'Hopital's Rule

L'Hopital's Rule states that if the limit of a ratio results in \( \frac{0}{0} \), we can take the partial derivatives of the numerator and denominator with respect to one of the variables, and then take the limit of the resulting expression.

Let’s focus on differentiating with respect to \( y \), since the expression seems to depend on \( y \) prominently.

The numerator is:

\[
f(y, x, z) = y - \sqrt{yx} - \sqrt{yz} + \sqrt{zx}.
\]

Differentiating the numerator with respect to \( y \) gives:

\[
\frac{\partial}{\partial y} \left( y - \sqrt{yx} - \sqrt{yz} + \sqrt{zx} \right) = 1 - \frac{1}{2\sqrt{yx}} \cdot x - \frac{1}{2\sqrt{yz}} \cdot z.
\]

Similarly, the denominator is:

\[
g(y, x, z) = y - \sqrt{yx} + \sqrt{yz} - \sqrt{zx},
\]

and its partial derivative with respect to \( y \) is:

\[
\frac{\partial}{\partial y} \left( y - \sqrt{yx} + \sqrt{yz} - \sqrt{zx} \right) = 1 - \frac{1}{2\sqrt{yx}} \cdot x + \frac{1}{2\sqrt{yz}} \cdot z.
\]

Now substitute \( x = 1 \), \( y = 1 \), and \( z = 4 \) into these expressions.

For the numerator:

\[
1 - \frac{1}{2\sqrt{1 \cdot 1}} \cdot 1 - \frac{1}{2\sqrt{1 \cdot 4}} \cdot 4 = 1 - \frac{1}{2} - \frac{4}{4} = 1 - \frac{1}{2} - 1 = -\frac{1}{2}.
\]

For the denominator:

\[
1 - \frac{1}{2\sqrt{1 \cdot 1}} \cdot 1 + \frac{1}{2\sqrt{1 \cdot 4}} \cdot 4 = 1 - \frac{1}{2} + \frac{4}{4} = 1 - \frac{1}{2} + 1 = \frac{3}{2}.
\]

Thus, the limit becomes:

\[
\lim_{(x, y, z) \to (1, 1, 4)} \frac{-\frac{1}{2}}{\frac{3}{2}} = \frac{-\frac{1}{2}}{\frac{3}{2}} = -\frac{1}{3}.
\]

### Final Answer:
The value of the limit is:

\[
\boxed{-\frac{1}{3}}.
\]

Would you like more details or have any questions?

---

Here are five related questions to explore:
1. What conditions lead to the application of L'Hopital's Rule?
2. How do partial derivatives influence multivariable limits?
3. Can L'Hopital's Rule be applied to any indeterminate form?
4. How does the behavior of square roots affect limit evaluation?
5. What other methods besides L'Hopital's Rule can handle indeterminate limits?

**Tip:** When encountering indeterminate forms, always check if simplifying the expression before applying advanced techniques is possible.

Evaluate the following limit. lim (x, y, z) → (1, 1, 4) (y - √(yx) - √(yz) + √(zx)) / (y - √(yx) + √(yz) - √(zx))

Learn how to evaluate the limit lim (x, y, z) → (1, 1, 4) of a complex multivariable expression using L'Hopital's Rule. This step-by-step solution covers key calculus concepts like indeterminate forms and partial derivatives. Ideal for advanced high school or undergraduate students.

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