Problem 11 - Calculus for Biology and Medicine

Question

In Problems 1-14, use the properties of limits to calculate the following limits: $$ \lim _{(x, y) \rightarrow(0,1)} \frac{2 x y-3}{x^{2}+y^{2}+1} $$

Step-by-Step Solution

Answer

The limit is \(-\frac{3}{2}\).

Step 1: Direct Substitution

First, try substituting the values of \(x\) and \(y\) directly into the limit function to check for indeterminate forms. Plug in \(x = 0\) and \(y = 1\): \(\frac{2(0)(1) - 3}{0^2 + 1^2 + 1} = \frac{-3}{2}\)This substitution does not lead to an indeterminate form, thus the limit can be directly evaluated.

Step 2: Evaluate the Limit

Since the substitution from Step 1 didn't lead to any form like \( \frac{0}{0} \) or other indeterminates, the limit can be evaluated directly as:\[\lim _{(x, y) \rightarrow(0,1)} \frac{2xy-3}{x^2+y^2+1} = \frac{-3}{2}\]

Key Concepts

Direct Substitution

Direct substitution is often the first method we try when evaluating limits. It involves plugging in the values toward which the variables are approaching directly into the function. This is a straightforward approach as long as it doesn't lead to any complications like division by zero or undefined expressions.

In our given problem, we substitute \(x = 0\) and \(y = 1\) directly into the expression \(\frac{2xy-3}{x^2+y^2+1}\). By doing so, we immediately find:

• Numerator: \(2(0)(1) - 3 = -3\)
• Denominator: \(0^2 + 1^2 + 1 = 2\)

Thus, the expression simplifies to \(\frac{-3}{2}\).

Using direct substitution is quick and efficient if it does not result in indeterminate forms. If it does result in an indeterminate form, then additional techniques for evaluating limits might be required.

Indeterminate Forms

Indeterminate forms occur when a direct substitution in a limit leads to expressions such as \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(0 \times \infty\), etc. These forms do not provide useful information about the actual limit and require further exploration to resolve.

In the exercise at hand, after substituting \(x = 0\) and \(y = 1\), we arrived at \(\frac{-3}{2}\) directly. This means that the limit does not result in an indeterminate form.
Knowing whether your answer results in an indeterminate form is critical because it tells you whether you need to apply other techniques, such as L'Hôpital's rule, algebraic manipulation, or geometric insights, to find the limit.

Evaluating Limits

Evaluating limits involves determining what value a function approaches as the variables within it get infinitely close to a certain point. This is essential in calculus because it helps us understand the behavior of functions at points where they might be undefined or at points of interest.

The primary goal when evaluating limits is to simplify the function such that no indeterminate forms arise. In our example, after successfully conducting direct substitution without encountering indeterminates, the limit was straightforwardly found to be \(\frac{-3}{2}\).

• Proper substitution: Ensures correct input to the function.
• Avoiding indeterminate forms: Confirms the validity of the approach.

This systematic process leads us to the correct evaluation of the limit without any additional complex calculation.