Finding the Simultaneous Limit of y sin(1/x)

The question asks for the simultaneous limit of the function $f(x,y) = y \sin(1/x)$. When evaluating simultaneous limits for functions of two variables, we are typically interested in the limit as $(x,y)$ approaches a specific point $(a,b)$. Given the form of the function, especially the $1/x$ term, the most common and interesting point to consider for the limit where $1/x$ becomes problematic is the origin, $(0,0)$. Let's find the limit as $(x,y) \to (0,0)$.

Understanding Simultaneous Limits

A simultaneous limit exists if the function approaches a single value regardless of the path taken by $(x,y)$ towards the limit point $(a,b)$. For the limit $\lim_{(x,y)\to(a,b)} f(x,y)$ to exist and be equal to $L$, for every $\epsilon > 0$, there must exist a $\delta > 0$ such that if $0 < \sqrt{(x-a)^2 + (y-b)^2} < \delta$, then $|f(x,y) - L| < \epsilon$.

Evaluating the Limit of y sin(1/x) as (x,y) → (0,0)

The function is $f(x,y) = y \sin(1/x)$. As $(x,y) \to (0,0)$, $y \to 0$. However, the term $\sin(1/x)$ behaves erratically as $x \to 0$. As $x$ approaches 0, $1/x$ goes to infinity (or minus infinity), and $\sin(1/x)$ oscillates rapidly between -1 and 1. This means $\lim_{x \to 0} \sin(1/x)$ does not exist.

Despite the non-existence of the limit of $\sin(1/x)$ as $x \to 0$, the presence of the $y$ term, which goes to 0 as $(x,y) \to (0,0)$, suggests we might be able to use the Squeeze Theorem (also known as the Sandwich Theorem).

Applying the Squeeze Theorem

We know that for any value $u$, the sine function is bounded:

$-1 \le \sin(u) \le 1$

Let $u = 1/x$. For $x \neq 0$, we have:

$-1 \le \sin(1/x) \le 1$

Now, let's multiply this inequality by $y$. We need to be careful depending on the sign of $y$. A simpler approach is to consider the absolute value of the function:

$|f(x,y)| = |y \sin(1/x)| = |y| |\sin(1/x)|$

Since $-1 \le \sin(1/x) \le 1$, the absolute value is bounded by 1:

$|\sin(1/x)| \le 1$ for $x \neq 0$.

Multiplying by $|y|$ (which is always non-negative), we get:

$|y| |\sin(1/x)| \le |y| \cdot 1$

$|y \sin(1/x)| \le |y|$

This inequality is equivalent to:

$-|y| \le y \sin(1/x) \le |y|$

Now, let's consider the limits of the bounding functions as $(x,y) \to (0,0)$.

• $\lim_{(x,y)\to(0,0)} -|y|$: As $(x,y) \to (0,0)$, $y \to 0$, so $|y| \to 0$. Thus, $\lim_{(x,y)\to(0,0)} -|y| = 0$.
• $\lim_{(x,y)\to(0,0)} |y|$: As $(x,y) \to (0,0)$, $y \to 0$, so $|y| \to 0$. Thus, $\lim_{(x,y)\to(0,0)} |y| = 0$.

Since the function $y \sin(1/x)$ is "squeezed" between two functions, $-|y|$ and $|y|$, both of which approach 0 as $(x,y) \to (0,0)$, by the Squeeze Theorem, the limit of $y \sin(1/x)$ as $(x,y) \to (0,0)$ must also be 0.

$\lim_{(x,y)\to(0,0)} y \sin(1/x) = 0$

Conclusion

The simultaneous limit of the function $y \sin(1/x)$ as $(x,y) \to (0,0)$ is 0.

Revision Table: Key Concepts for Simultaneous Limits

Additional Information: Understanding the Behavior Near (0,0)

Consider the behavior of $y \sin(1/x)$ near $(0,0)$.

• If we approach along the y-axis (where $x=0$), the function is not defined.
• If we approach along any line $y = mx$, the function becomes $mx \sin(1/x)$. As $x \to 0$, $mx \to 0$, and $\sin(1/x)$ is bounded. The product of a term going to 0 and a bounded term goes to 0. So the limit along $y=mx$ is 0.
• If we approach along the x-axis (where $y=0$), the function is $0 \cdot \sin(1/x) = 0$ (for $x \neq 0$). As $x \to 0$, the function is 0, so the limit along the x-axis is 0.

While checking paths can suggest a limit exists, it doesn't prove it. The Squeeze Theorem provides the rigorous proof that the simultaneous limit is indeed 0. The key is that the factor $y$ forces the function value towards 0 as $(x,y)$ approaches $(0,0)$, regardless of the oscillatory behavior of $\sin(1/x)$.