Problem 82 - Calculus Volume 3

Question

For the following exercises, use algebraic techniques to evaluate the limit. $$\quad \lim _{(x, y) \rightarrow(0,0)} \frac{x^{3}-y^{3}}{x-y}$$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Understand the Limit

The limit we need to evaluate is \( \lim_{(x, y) \to (0,0)} \frac{x^3 - y^3}{x-y} \). Since the limit approaches \((0, 0)\), the expression \(\frac{x^3 - y^3}{x-y}\) becomes an indeterminate form \(\frac{0}{0}\). Therefore, we need to simplify this expression to find the limit.

Step 2: Apply Algebraic Simplification

Notice that \(x^3 - y^3\) is a difference of cubes. Recall the formula for the difference of cubes: \[x^3 - y^3 = (x-y)(x^2 + xy + y^2)\].Plug this back into the fraction: \[\frac{x^3 - y^3}{x-y} = \frac{(x-y)(x^2 + xy + y^2)}{x-y}.\]

Step 3: Cancel the Common Factor

Since \(xeq y\) except at the limit point, we can cancel out the \((x-y)\) term in both the numerator and the denominator:\[\frac{(x-y)(x^2 + xy + y^2)}{x-y} = x^2 + xy + y^2.\]

Step 4: Evaluate the Limit of the Simplified Expression

Now, find the limit of \(x^2 + xy + y^2\) as \((x, y)\) approaches \((0, 0)\). Substitute \(x = 0\) and \(y = 0\) into the expression:\[\lim_{(x, y) \to (0,0)} x^2 + xy + y^2 = 0^2 + 0 \cdot 0 + 0^2 = 0.\]

Step 5: Conclusion

The function simplifies to a polynomial for which we directly calculated the limit as both variables approach zero. The limit of the original expression is then 0.

Key Concepts

Algebraic Techniques for Limits

When evaluating limits in multivariable calculus, algebraic techniques are powerful tools to simplify complex expressions. A limit can become much more manageable by rewriting parts of a function. Here are steps to apply these techniques successfully:

• Identify any indeterminate forms, typically expressed as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).
• Use algebraic manipulation to rewrite the expression, often by factoring or expanding parts of the function.
• Substitute or cancel terms to make the expression easier to evaluate.

Break down the expression to its core components. This might involve using different algebraic formulas or identities. Once simplified, limits can typically be evaluated directly by substituting the limit point, making complex problems much more approachable.

Indeterminate Forms

In the context of limits, an expression is said to be an indeterminate form when its limit cannot be directly evaluated. One common example is \(\frac{0}{0}\), which arises when both the numerator and denominator approach zero as a variable approaches a certain point.

Indeterminate forms require additional work to resolve:

• Inspect the original expression for possible factorizations or function identities that might simplify it.
• Apply algebraic transformations, such as factoring or multiplying by conjugates, to rewrite the expression.
• Cancel out terms or use special limit laws designed to handle these situations.

Handling indeterminate forms is all about transforming the expression into something manageable, from which the limit can be directly evaluated after resolving indeterminacies.

Difference of Cubes Formula

The difference of cubes is a specific algebraic identity that helps simplify expressions where variables are raised to the third power. It is especially handy in simplifying expressions that initially appear as indeterminate forms.The identity is expressed as:
\[ x^3 - y^3 = (x-y)(x^2 + xy + y^2) \]
This formula shows us how to break down a complex cubic expression into simpler polynomial terms. Here is how it aids in solving limits:

• It provides a way to factor and reduce the expression into non-indeterminate form.
• By factoring a difference of cubes, you simplify the numerator, revealing common terms with the denominator.
• When the factor \((x-y)\) is cancelled, it often resolves the \(\frac{0}{0}\) indeterminacy at points where \(x=y\).

Understanding and applying the difference of cubes formula is crucial for simplifying and evaluating multivariable limits effectively.