Problem 80 - Calculus Early Transcendentals

Question

Evaluate the following limits. $$\lim _{(x, y) \rightarrow(0, \pi / 2)} \frac{1-\cos x y}{4 x^{2} y^{3}}$$

Step-by-Step Solution

Answer

The limit as (x, y) approaches (0, π/2) of the function f(x,y) = (1- cos(xy))/(4x²y³) is 0.

Step 1: Verify the conditions for L'Hopital's rule

For L'Hopital's rule to be applied in this case, the following conditions must be met: 1. The limit has to be in the form 0/0 or ±∞/±∞ as \((x, y) \rightarrow (0, \pi/2)\). 2. Partial derivatives of the numerator and denominator with respect to both variables exist and are continuous in the neighborhood of the point \((0, \pi/2)\). Let's check if these conditions are met. As \((x, y) \rightarrow (0, \pi/2)\), the numerator becomes: $$1 - \cos(0 \cdot \pi/2) = 1 - \cos(0) = 1 - 1 = 0$$ The denominator becomes: $$4 \cdot 0^2 \cdot (\pi/2)^3 = 0$$ Since we have the form 0/0, the first condition is met. Now let's check if partial derivatives exist and are continuous in the neighborhood of \((0, \pi/2)\).

Step 2: Find partial derivatives of the numerator and denominator

We will find the first partial derivatives of the numerator and denominator with respect to both x and y. For the numerator \(u(x, y) = 1 - \cos(xy)\): $$\frac{\partial u}{\partial x} = y \sin(xy)$$ $$\frac{\partial u}{\partial y} = x \sin(xy)$$ For the denominator \(v(x, y) = 4x^2y^3\): $$\frac{\partial v}{\partial x} = 8xy^3$$ $$\frac{\partial v}{\partial y} = 12x^2y^2$$ All these partial derivatives are continuous in the neighborhood of \((0, \pi/2)\), so the second condition is met. We can now apply L'Hopital's rule.

Step 3: Apply L'Hopital's rule

L'Hopital's rule for multivariable limits states that if the conditions above are met, then: $$\lim_{(x, y) \rightarrow (0, \pi/2)} \frac{1-\cos xy}{4x^2y^3} = \lim_{(x, y) \rightarrow (0, \pi/2)} \frac{\frac{\partial(u)}{\partial x} \frac{\partial(v)}{\partial y} - \frac{\partial(u)}{\partial y} \frac{\partial(v)}{\partial x}}{(\frac{\partial(v)}{\partial x})^2 + (\frac{\partial(v)}{\partial y})^2}$$ Applying the derivatives we found in step 2: $$\lim_{(x, y) \rightarrow (0, \pi/2)} \frac{(y \sin(xy))(12x^2y^2) - (x \sin(xy))(8xy^3)}{(8xy^3)^2 + (12x^2y^2)^2}$$

Step 4: Simplify the expression

Let's simplify the expression inside the limit: $$\lim_{(x, y) \rightarrow (0, \pi/2)} \frac{12x^2y^3 \sin(xy) - 8x^2y^3 \sin(xy)}{64x^4y^6 + 144x^4y^4}$$ Factor out common terms: $$\lim_{(x, y) \rightarrow (0, \pi/2)} \frac{4x^2y^3\sin(xy)(3-2)}{16x^4y^4(y^2+9)}$$ Reduce the expression: $$\lim_{(x, y) \rightarrow (0, \pi/2)} \frac{y\sin(xy)}{4(y^2+9)}$$

Step 5: Evaluate the limit

Now, we can evaluate the limit as \((x, y) \rightarrow (0, \pi/2)\): $$\lim_{(x, y) \rightarrow (0, \pi/2)} \frac{\pi/2\sin(0)}{4((\pi/2)^2+9)} = \frac{0}{4((\pi/2)^2+9)} = 0$$ So, the limit as \((x, y) \rightarrow (0, \pi/2)\) is: $$\lim_{(x, y) \rightarrow (0, \pi/2)} \frac{1-\cos xy}{4x^2y^3} = 0$$