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The image contains a limit problem. It is written as follows:

\[
\lim_{(x,y) \to (0,0)} \frac{x^4 + 5y^3}{x^2 + y^2}
\]

Let's evaluate the limit step by step.

### Step 1: Analyze the function

The function is:
\[
f(x, y) = \frac{x^4 + 5y^3}{x^2 + y^2}
\]

We need to determine whether the limit exists as \((x, y) \to (0, 0)\).

### Step 2: Try different paths to the origin

One common approach to evaluate two-variable limits is to try different paths to the origin and check if the limit value is the same for all these paths.

#### Path 1: Along the \(x\)-axis (\(y = 0\))
\[
f(x, 0) = \frac{x^4 + 5(0)^3}{x^2 + 0^2} = \frac{x^4}{x^2} = x^2
\]
As \(x \to 0\), we have:
\[
f(x, 0) = 0
\]

#### Path 2: Along the \(y\)-axis (\(x = 0\))
\[
f(0, y) = \frac{(0)^4 + 5y^3}{(0)^2 + y^2} = \frac{5y^3}{y^2} = 5y
\]
As \(y \to 0\), we have:
\[
f(0, y) = 0
\]

#### Path 3: Along the line \(y = x\)
\[
f(x, x) = \frac{x^4 + 5x^3}{x^2 + x^2} = \frac{x^4 + 5x^3}{2x^2} = \frac{x^2(x + 5)}{2}
\]
As \(x \to 0\), we have:
\[
f(x, x) = 0
\]

### Step 3: Polar coordinates

To further verify, let's convert the function into polar coordinates, where \(x = r\cos\theta\) and \(y = r\sin\theta\).

\[
f(r, \theta) = \frac{(r\cos\theta)^4 + 5(r\sin\theta)^3}{(r\cos\theta)^2 + (r\sin\theta)^2} = \frac{r^4\cos^4\theta + 5r^3\sin^3\theta}{r^2(\cos^2\theta + \sin^2\theta)} = \frac{r^4\cos^4\theta + 5r^3\sin^3\theta}{r^2}
\]
\[
= r^2\cos^4\theta + 5r\sin^3\theta
\]
As \(r \to 0\), both terms go to 0, regardless of the angle \(\theta\).

### Conclusion:

Since the limit is 0 along all the paths (including polar coordinates), the limit exists and is:

\[
\lim_{(x,y) \to (0,0)} \frac{x^4 + 5y^3}{x^2 + y^2} = 0
\]

Would you like a detailed explanation for any specific step or have any further questions?

Here are 5 related questions you might want to explore:
1. What other methods can be used to check the existence of multivariable limits?
2. How does polar coordinate substitution help in evaluating limits?
3. What is the difference between approaching limits along straight lines versus curved paths?
4. How does one determine if a limit does not exist for multivariable functions?
5. What are indeterminate forms, and how do they affect limit evaluation?

**Tip:** For multivariable limits, always check different paths and convert to polar coordinates when in doubt—it often simplifies the process.

Evaluate the limit: lim (x,y) -> (0,0) of (x^4 + 5y^3) / (x^2 + y^2).

Learn how to evaluate the multivariable limit lim (x,y) -> (0,0) of the function (x^4 + 5y^3) / (x^2 + y^2) using path testing and polar coordinates. Suitable for university-level calculus students.

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