Problem 4 - Calculus for Biology and Medicine

Question

In Problems 1-14, use the properties of limits to calculate the following limits: $$ \lim _{(x, y) \rightarrow(1,-2)}\left(2 x^{3}-3 y\right)(x y-2) $$

Step-by-Step Solution

Answer

The limit is -32.

Step 1: Identify the Components

Identify the two main components of the given expression: \( (2x^3 - 3y) \) and \( (xy - 2) \). We will separately find the limits of these components as \( (x, y) \to (1, -2) \).

Step 2: Calculate the Limit of the First Component

First, calculate the limit of \( 2x^3 - 3y \). Substituting \( x = 1 \) and \( y = -2 \), we get: \( 2(1)^3 - 3(-2) = 2 - (-6) = 8 \). Therefore, \( \lim_{(x, y) \to (1, -2)}(2x^3 - 3y) = 8 \).

Step 3: Calculate the Limit of the Second Component

Next, calculate the limit of \( xy - 2 \). Substituting \( x = 1 \) and \( y = -2 \), we get: \( (1)(-2) - 2 = -2 - 2 = -4 \). Therefore, \( \lim_{(x, y) \to (1, -2)}(xy - 2) = -4 \).

Step 4: Combine the Limits of Both Components

Use the property of limits that states the limit of the product is the product of the limits: \( \lim_{(x, y) \to (1, -2)} (2x^3 - 3y)(xy - 2) = 8 \times (-4) = -32 \).

Key Concepts

Properties of Limits

Understanding the properties of limits is an essential part of working with limit calculations in calculus. Among the properties, one of the most useful is the limit of a product property. This states that the limit of a product of functions is equal to the product of their limits, provided each limit exists independently. In solving multivariable limit problems, these properties simplify the evaluation process significantly. For instance, in our initial problem, we used this property to combine the limits of two separate components of the expression. When evaluating \( \lim_{(x, y) \to (1, -2)} \), we computed the limits of \( 2x^3 - 3y \) and \( xy - 2 \) independently. Since both these limits exist as we approach \( (1, -2) \), we can confidently multiply their results—8 and -4 respectively—to find the overall limit.

Multivariable Calculus

In calculus, extending concepts from single variable to multivariable functions introduces a few complexities. Multivariable calculus deals with functions of two or more variables, and this requires evaluating limits as a point approaches a particular location in a plane (or even higher dimensions). Unlike single-variable limits, multivariable limits must be approached along many paths as we near a specific point.
When finding limits in multivariable calculus, it’s crucial to consider approaches from different directions and paths. However, many problems, especially when neatly factored like this one, allow us to use the properties of limits effectively without having to check every possible path. Still, understanding how functions behave in multiple dimensions helps resolve potential inconsistencies across different paths, ensuring that our calculated limit is indeed correct.

Limit Calculation

Calculating limits can sometimes feel daunting, but by breaking down the process, it becomes more manageable. Let's consider the given function: \((2x^3 - 3y)(xy - 2)\). The task is to compute its limit as \((x, y)\) approaches \((1, -2)\).

• **First Component:** For \( 2x^3 - 3y \), substitute the known point \( (1, -2) \). This simplifies the expression through straightforward arithmetic to find a limit of 8.
• **Second Component:** Similarly, for \( xy - 2 \), substitute the values \( (1, -2) \) to get -4.
• **Combine the Results:** Once we've calculated these individual limits, we use the product property to combine them, resulting in the final limit of \(-32\).

This step-by-step breakdown not only clarifies the calculation process but also shows how properties of limits help in solving more complex problems efficiently.