The image you provided contains two limit problems. Let's solve each of them step-by-step.

Problem 4:

$\lim_{(x,y) \to (1,2)} \frac{2xy}{x^2 + y^2}$

Step 1: Plug in the limit point (x, y) = (1, 2)

Substitute $x = 1$ and $y = 2$ into the expression: $\frac{2xy}{x^2 + y^2} = \frac{2(1)(2)}{1^2 + 2^2}$

Step 2: Simplify the expression

Calculate both the numerator and the denominator: $\text{Numerator} = 2 \times 1 \times 2 = 4$ $\text{Denominator} = 1^2 + 2^2 = 1 + 4 = 5$

Thus, the expression simplifies to: $\frac{4}{5}$

Final answer for Problem 4: $\boxed{\frac{4}{5}}$

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Problem 5:

$\lim_{(x, y) \to (0, 1)} \left[e^{x^2 + x} \ln(ey^2)\right]$

Step 1: Simplify the expression

First, recall the property of logarithms: $\ln(e) = 1$. This simplifies $\ln(ey^2)$ as: $\ln(ey^2) = \ln(e) + \ln(y^2) = 1 + 2\ln(y)$ Thus, the limit becomes: $\lim_{(x, y) \to (0, 1)} e^{x^2 + x} \left(1 + 2\ln(y)\right)$

Step 2: Evaluate the limit

Now, plug in $x = 0$ and $y = 1$:

1. For $e^{x^2 + x}$, when $x = 0$, we get: $e^{0^2 + 0} = e^0 = 1$

2. For $1 + 2\ln(y)$, when $y = 1$, we get: $1 + 2\ln(1) = 1 + 2(0) = 1$

Thus, the entire expression becomes: $1 \times 1 = 1$

Final answer for Problem 5: $\boxed{1}$

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Do you have any questions or would you like a more detailed explanation of any step?

Related Questions:

1. How do you handle limits approaching infinity for multi-variable functions?
2. Can we use polar coordinates to solve limits for certain multi-variable functions?
3. What are some common indeterminate forms encountered in multi-variable limits?
4. How does L'Hôpital's rule extend to multi-variable calculus?
5. How can you evaluate limits involving trigonometric functions in two variables?

Tip:

Always check if direct substitution works first when solving limits. It often saves time and effort before considering more advanced techniques like L'Hôpital's rule or using polar coordinates.