Question

Finding a Limit In Exercises \(41-46,\) write a simpler function that agrees with the given function at all but one point. Then find the limit of the function. Use a graphing utility to confirm your result. $$ \lim _{x \rightarrow-1} \frac{x^{3}+1}{x+1} $$

Short answer

The limit of the function as \(x\) approaches -1 is 3.

Step-by-step solution

Simplify the function

We begin by simplifying the function. The numerator can be factored using the difference of cubes formula, which states that \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\). So, we have: \n\(x^3 + 1 = (x+1)(x^2 - x + 1)\). \nWe can then replace the numerator of the original function with its factored form: \nThe function becomes \(\frac{(x+1)(x^2 - x + 1)}{x+1}\).

Cancel out common factor

Next, we cancel out the \(x+1\) from the numerator and the denominator. This removes the point where the function is undefined. So we get a simpler function: \(x^2 - x + 1\).

Compute the limit

Finally, we substitute the value \(x = -1\) into the simplified function, as the function is now defined for \(x = -1\). This yields: \n\[\lim _{x \rightarrow-1} x^2 - x + 1 = (-1)^2 - (-1) + 1 = 1 + 1 + 1 = 3.\]

Key concepts

Difference of Cubes

When working with polynomials, you might come across a form known as the "difference of cubes." This formula is helpful because it allows you to express a sum or difference of cubes in a factored form. For the sum of cubes, like in our exercise, we use the identity:\[ a^3 + b^3 = (a+b)(a^2 - ab + b^2) \]

In the expression \(x^3 + 1\), this is a sum of cubes where \(a = x\) and \(b = 1\). Factoring it gives us \((x+1)(x^2 - x + 1)\). This step is crucial because it helps simplify complex expressions into simpler components.

Using these formulas, challenges involving cubic polynomials become more manageable, laying a clear path toward further simplifications or evaluations.

Simplifying Functions

In calculus, simplifying a function is a key step to make analyses, such as finding limits, easier. The goal is to transform the original expression into a simpler one without changing its essential properties. By shedding unnecessary complexity, we can examine the function's behavior more clearly.

Starting with the equation \(\frac{x^3+1}{x+1}\), we first factor the numerator using the difference of cubes, resulting in \(\frac{(x+1)(x^2 - x + 1)}{x+1}\). Now, the function appears convoluted, but its behavior is exactly the same for every \(x\) except the problematic point that initially caused the function to be undefined.

Simplifying the function not only clarifies the structure of the problem but also reveals underlying themes or patterns in calculus that wouldn't be apparent beforehand.

Factoring

Factoring is a mathematical process that involves breaking down an expression into a product of its simpler divisions or factors. It provides a critical step for solving and simplifying polynomial equations.

In our exercise, the polynomial \(x^3+1\) is factored into \((x+1)(x^2-x+1)\) using the sum of cubes formula. Factoring exposes the roots or solutions of the polynomial by expressing it in simpler, multipliable components.

This restructuring is particularly useful here, as it allows us to later cancel common factors, simplifying expressions, and facilitating limit evaluation. Like turning a complex cake into its basic ingredients, factoring makes each part of the expression easier to handle, interpret, and solve.

Canceling Common Factors

Once you have factored a polynomial expression, the next step often involves canceling any common factors found in the numerator and denominator. This is especially helpful when the fractions initially result in undefined values at certain points, like \(x = -1\) in our case.

After factoring \(\frac{(x+1)(x^2-x+1)}{x+1}\), we see that \(x+1\) is common to both the numerator and denominator. Canceling these allows us to redefine the function as \(x^2-x+1\), getting rid of the problematic undefined point.

• This simplification is crucial as it reveals a continuous and well-behaved function in the vicinity of the point previously causing issues.
• This new function, \(x^2-x+1\), can now safely be evaluated for \(x = -1\) to determine the limit.

Canceling common factors enhances the function's clarity and sets the stage for straightforward evaluation of its limit.