It looks like you're asking about the behavior of the function

$f(x, y) = \frac{\sin (\pi x)}{2x + y}$

as $(x, y) \to (+\infty, +\infty)$. Let's analyze its limit.

Step 1: Numerator Analysis

The numerator is $\sin (\pi x)$, which oscillates between $-1$ and $1$. This means it remains bounded.

Step 2: Denominator Analysis

The denominator, $2x + y$, tends to $+\infty$ as both $x$ and $y$ go to $+\infty$. Specifically:

• If $y = 0$, then $2x + y = 2x$, which grows to $+\infty$.
• If $y = x$, then $2x + y = 3x$, which also grows to $+\infty$.
• In general, for any positive $y$, the denominator keeps increasing.

Step 3: Limit Calculation

Since the numerator is always bounded ($-1 \leq \sin(\pi x) \leq 1$) and the denominator grows arbitrarily large, the fraction approaches zero:

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

Thus, the function tends to 0 in this limit.

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Would you like a more detailed explanation or a different approach?

Related Questions:

1. How does the function behave if $x \to +\infty$ and $y$ remains constant?
2. What happens if we replace $\sin (\pi x)$ with another periodic function?
3. How does the function behave in the limit $(x, y) \to (-\infty, -\infty)$?
4. What if the denominator were $2x - y$ instead of $2x + y$?
5. How would you evaluate the limit using an $\epsilon-\delta$ approach?

Tip:

When dealing with limits involving oscillatory functions, check if the denominator dominates (grows indefinitely). If it does, the fraction often tends to zero.