lim (x,y) 0,0,) (sqrt(x) + sqrt(y))/(x + 2sqrt(x) - y + 2sqrt(y))

We are tasked with evaluating the limit:

\[
\lim_{(x, y) \to (0, 0)} \frac{\sqrt{x} + \sqrt{y}}{x + 2\sqrt{x} - y + 2\sqrt{y}}
\]

### Step-by-Step Solution

1. **Substitute \(x = 0\) and \(y = 0\) directly:**

   First, let's substitute \(x = 0\) and \(y = 0\) into the expression to check if we get a defined result:
   \[
   \frac{\sqrt{0} + \sqrt{0}}{0 + 2\sqrt{0} - 0 + 2\sqrt{0}} = \frac{0 + 0}{0 + 0 - 0 + 0} = \frac{0}{0}
   \]
   This results in an indeterminate form \(\frac{0}{0}\), so we must investigate further using other methods like **polar coordinates** or **path analysis**.

2. **Approach along the path \( x = y \):**

   Substituting \(x = y\) into the original expression:
   \[
   \frac{\sqrt{x} + \sqrt{x}}{x + 2\sqrt{x} - x + 2\sqrt{x}} = \frac{2\sqrt{x}}{2\sqrt{x}} = 1
   \]
   Along the path \(x = y\), the limit appears to be 1.

3. **Approach along the path \( y = 0 \):**

   Substituting \(y = 0\) into the original expression:
   \[
   \frac{\sqrt{x} + \sqrt{0}}{x + 2\sqrt{x} - 0 + 2\sqrt{0}} = \frac{\sqrt{x}}{x + 2\sqrt{x}} = \frac{1}{\frac{x}{\sqrt{x}} + 2} = \frac{1}{\sqrt{x} + 2}
   \]
   As \(x \to 0\), the expression tends to:
   \[
   \frac{1}{2}
   \]
   Along the path \(y = 0\), the limit is \(\frac{1}{2}\).

4. **Approach along the path \(x = 0\):**

   Substituting \(x = 0\) into the original expression:
   \[
   \frac{\sqrt{0} + \sqrt{y}}{0 + 2\sqrt{0} - y + 2\sqrt{y}} = \frac{\sqrt{y}}{- y + 2\sqrt{y}} = \frac{1}{-\frac{y}{\sqrt{y}} + 2} = \frac{1}{- \sqrt{y} + 2}
   \]
   As \(y \to 0\), the expression tends to:
   \[
   \frac{1}{2}
   \]
   Along the path \(x = 0\), the limit is \(\frac{1}{2}\).

### Conclusion

Since the limit along different paths yields different results (1 along \(x = y\) and \(\frac{1}{2}\) along \(x = 0\) or \(y = 0\)), the limit does **not exist**.

### Would you like further clarification or details?

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### Related Questions:

1. What does it mean for a limit to be indeterminate?
2. Can polar coordinates simplify similar limit problems?
3. How do path-dependent limits help in determining if a limit exists?
4. What are other common methods for evaluating two-variable limits?
5. How would the limit change if a different path is chosen?

### Tip:
For multivariable limits, always check multiple paths. If different paths give different limits, the overall limit does not exist.

lim (x,y) -> (0,0) (sqrt(x) + sqrt(y))/(x + 2sqrt(x) - y + 2sqrt(y))

Learn how to evaluate the limit of (sqrt(x) + sqrt(y))/(x + 2sqrt(x) - y + 2sqrt(y)) as (x, y) approaches (0, 0). This solution explores path-dependent limits, indeterminate forms, and the conditions under which a limit does not exist, suitable for college-level multivariable calculus students.

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