	lim┬((x,y)→(0,0) )⁡〖(x^3+ y^2)/(√(〖x^2+y〗^2+4)   - 2)〗

To evaluate the limit

\[
\lim_{(x,y) \to (0,0)} \frac{x^3 + y^2}{\sqrt{x^2 + y^2 + 4} - 2},
\]

we can start by simplifying the expression.

### Step 1: Consider the denominator
The expression in the denominator is of the form \(\sqrt{A} - B\), where \(A = x^2 + y^2 + 4\) and \(B = 2\). A common technique to simplify such expressions is to multiply the numerator and the denominator by the conjugate of the denominator, i.e., \(\sqrt{x^2 + y^2 + 4} + 2\).

### Step 2: Multiply by the conjugate
Multiply the numerator and the denominator by \(\sqrt{x^2 + y^2 + 4} + 2\):

\[
\frac{(x^3 + y^2)(\sqrt{x^2 + y^2 + 4} + 2)}{(\sqrt{x^2 + y^2 + 4} - 2)(\sqrt{x^2 + y^2 + 4} + 2)}.
\]

### Step 3: Simplify the expression
The denominator becomes a difference of squares:

\[
(\sqrt{x^2 + y^2 + 4})^2 - 2^2 = (x^2 + y^2 + 4) - 4 = x^2 + y^2.
\]

So, the expression simplifies to:

\[
\frac{(x^3 + y^2)(\sqrt{x^2 + y^2 + 4} + 2)}{x^2 + y^2}.
\]

### Step 4: Analyze the limit
Now, we need to evaluate the limit of this expression as \((x, y) \to (0,0)\).

1. **For the numerator**: As \((x, y) \to (0,0)\), \(x^3 \to 0\) and \(y^2 \to 0\), so the numerator \((x^3 + y^2)\) approaches 0.

2. **For the denominator**: As \((x, y) \to (0,0)\), \(x^2 + y^2 \to 0\), but the term \(\sqrt{x^2 + y^2 + 4} + 2\) approaches 4 (since \(\sqrt{4} = 2\)).

Thus, the expression simplifies to:

\[
\lim_{(x,y) \to (0,0)} \frac{0 \cdot (\text{something finite})}{0} = 0.
\]

Since the numerator tends to 0 faster than the denominator (which is nonzero but small), the limit of the entire expression is 0.

So the final answer is:

\[
\lim_{(x,y) \to (0,0)} \frac{x

Learn how to evaluate the limit of (x^3 + y^2) / (sqrt(x^2 + y^2 + 4) - 2) as (x, y) approaches (0,0). This problem involves advanced concepts in limits and multivariable calculus. Suitable for advanced college-level students.

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