We need to evaluate the limit:

$\lim_{(x,y) \to (+\infty, +\infty)} \frac{\sin (\pi x)}{2x + y}$

Step 1: Analyze the Numerator

The function $\sin(\pi x)$ oscillates between $-1$ and $1$ for all values of $x$. That means:

$-1 \leq \sin(\pi x) \leq 1$

Step 2: Analyze the Denominator

The denominator is:

$2x + y$

Since both $x \to +\infty$ and $y \to +\infty$, the denominator grows indefinitely. This means:

$|2x + y| \to +\infty$

Step 3: Apply the Squeeze Theorem

Since $-1 \leq \sin(\pi x) \leq 1$, we divide by $2x + y$:

$\frac{-1}{2x + y} \leq \frac{\sin(\pi x)}{2x + y} \leq \frac{1}{2x + y}$

Since $2x + y \to +\infty$, both bounds tend to $0$. By the Squeeze Theorem, we conclude:

$\lim_{(x,y) \to (+\infty, +\infty)} \frac{\sin (\pi x)}{2x + y} = 0$

Final Answer:

$\lim_{(x,y) \to (+\infty, +\infty)} \frac{\sin (\pi x)}{2x + y} = 0$

Would you like more details or another approach?

Related Questions:

1. What is the limit if $y$ approaches infinity while $x$ remains constant?
2. How does the limit change if the denominator is $x + y^2$?
3. What happens if we replace $\sin(\pi x)$ with $\cos(\pi y)$?
4. Can we use L'Hôpital’s Rule for this type of limit?
5. What if $x, y \to (-\infty, -\infty)$ instead?

Tip: When dealing with limits at infinity, checking the growth rate of the denominator compared to the numerator helps determine if the fraction tends to zero.