Problem 9 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The limit as \((x, y) \rightarrow (1, 0)\) is 1.

Step 1: Identify the Expression and Limits

We are given the limit \( \lim _{(x, y) \rightarrow(1,0)} \frac{x^{2}+y^{2}}{x^{2}-y^{2}} \). This means we need to evaluate \( \frac{x^{2}+y^{2}}{x^{2}-y^{2}} \) as \( x \) approaches 1 and \( y \) approaches 0.

Step 2: Substitute Limits into the Expression

Substitute \( x = 1 \) and \( y = 0 \) into the expression: \[ \frac{1^2 + 0^2}{1^2 - 0^2} = \frac{1}{1} \] The initial substitution shows that the expression simplifies directly to \( 1 \).

Step 3: Verify Special Cases for Continuity

Since both the numerator \( x^2 + y^2 \) and the denominator \( x^2 - y^2 \) are continuous at \((x, y) = (1, 0)\), and the denominator is not zero at \((1, 0)\), the limit is valid. Thus, our calculation doesn't face any divisions by zero or undefined behavior.

Key Concepts

Continuity

When discussing continuity, we are referring to a fundamental property of functions that ensures they behave predictably when inputs are altered ever so slightly. A function is considered continuous at a particular point if the limit of the function at that point is equal to the function's value at the same point.
For example, in our given problem, the functions in both the numerator and the denominator, specifically \(x^2 + y^2\) and \(x^2 - y^2\), are continuous at the point \(1,0\). This suggests that if you alter \(x\) or \(y\) slightly around \(1\) and \(0\) respectively, the expression's behavior around this point will not abruptly change.

• The continuity of a function helps in predicting its behavior in relation to limits.
• If both the numerator and denominator of a fraction are continuous, it usually avoids undefined behavior at the specified point of evaluation.

Clearly, due to continuity, evaluating the limit becomes much more straightforward.

Limits of Functions

Limits play a vital role in calculus, as they help describe how a function behaves as its input gets infinitely close to a certain value. Within our context, the limit allows us to examine what happens to the fraction \(\frac{x^2+y^2}{x^2-y^2}\) as \((x,y)\) approaches \((1,0)\).
Understanding limits of functions is crucial because it forms the base for defining derivatives and integrals, which are core to calculus. When evaluating a limit:

• Identify if direct substitution results in an undefined form like \(\frac{0}{0}\) or \(\frac{\text{non-zero}}{0}\).
• Evaluate whether the functions involved are continuous at the point and if the substitution results in a valid expression.

In our context, substituting directly into the expression gives \(\frac{1}{1} = 1\), indicating a straightforward solution with no undefined forms encountered.

Multivariable Calculus

Multivariable calculus extends single-variable concepts like limits and derivatives to functions with more than one input variable. It deals with topics such as partial derivatives, gradients, and evaluating functions like \(\frac{x^2+y^2}{x^2-y^2}\) over a plane.
When solving limits in multivariable calculus, such as the one given, it involves:

• Understanding how each variable contributes to the function.
• Evaluating limits by considering paths of approach in 2D space, ensuring the limit remains consistent regardless of how the point \((1,0)\) is approached.

Multivariable problems demand a deeper understanding because different paths or approaches could yield different results without assurance of continuity. However, with the continuity of both the numerator and the denominator, as in our example, multivariable limit calculations are simplified by direct substitution.