Question

Evaluate the following limits. $$\lim _{(x, y) \rightarrow(1,0)} \frac{\sin x y}{x y}$$

Short answer

Question: Evaluate the limit of the function $$f(x, y) = \frac{\sin xy}{xy}$$ as the point (x, y) approaches (1, 0). Answer: The limit of the function as the point (x, y) approaches (1, 0) is 1. Explanation: By analyzing the function along different paths that approach the point (1, 0), it is determined that the limit exists and is equal to 1 for all paths.

Step-by-step solution

Write down the given function

The given function is: $$f(x, y) = \frac{\sin xy}{xy}$$

Consider approaching the point (1,0) from various paths

To determine if the limit exists as \((x, y) \rightarrow (1,0)\), we will consider approaching the point from various paths. If the limit is the same for all paths, then the limit exists. Let's start with the path \(y = mx - m\), where \(m\) is a constant. This path represents lines that pass through the point (1, 0) with different slopes.

Substitute the path into the function

We now substitute the expression for \(y\) from the path (Step 2) into the given function: $$g(x, m) = \frac{\sin x(mx - m)}{x(mx - m)}$$

Simplify the function

We can simplify \(g(x, m)\): $$g(x, m) = \frac{\sin(mx^2 - mx)}{mx^2 - mx}$$

Determine the limit along the path

Now we evaluate the limit along the path as \(x \rightarrow 1\): $$\lim_{x \rightarrow 1} g(x, m) = \lim_{x \rightarrow 1} \frac{\sin(mx^2 - mx)}{mx^2 - mx}$$

Apply L'Hopital's rule

Since the limit involves an indeterminate form of the type \(\frac{0}{0}\), we can apply L'Hôpital's rule: $$\lim_{x \rightarrow 1} g(x, m) = \lim_{x \rightarrow 1} \frac{\frac{d}{dx}(\sin(mx^2 - mx))}{\frac{d}{dx}(mx^2 - mx)}$$

Differentiate the numerator and denominator

We have to differentiate both the numerator and denominator with respect to \(x\): $$\frac{d}{dx}(\sin(mx^2 - mx)) = \cos(mx^2 - mx)(2mx - m)$$ $$\frac{d}{dx}(mx^2 - mx) = 2mx - m$$

Compute the limit with the derivatives

Now we substitute the derivatives into the limit: $$\lim_{x \rightarrow 1} \frac{\cos(mx^2 - mx)(2mx - m)}{2mx - m}$$

Cancel the common factor

We can cancel the common factor of \(2mx - m\) from the numerator and denominator: $$\lim_{x \rightarrow 1} \cos(mx^2 - mx)$$

Evaluate the limit

Now we can evaluate the limit as \(x \rightarrow 1\): $$\lim_{x \rightarrow 1} \cos(mx^2 - mx) = \cos(m(1) - m) = \cos(0) = 1$$ Since the limit is the same for all paths, we can conclude that the limit exists and is equal to 1: $$\lim _{(x, y) \rightarrow(1,0)} \frac{\sin x y}{x y} = 1$$