Problem 66 - Calculus: Early Transcendentals

Question

Find the following limits or state that they do not exist. Assume \(a, b, c,\) and k are fixed real numbers. $$\lim _{x \rightarrow-1} g(x), \text { where } g(x)=\left\\{\begin{array}{ll}\frac{x^{2}-1}{x+1} & \text { if } x<-1 \\ -2 & \text { if } x \geq-1\end{array}\right.$$

Step-by-Step Solution

Answer

\(g(x) = \begin{cases} \frac{x^2-1}{x+1} & x<-1 \\ -2 & x\geq-1 \end{cases} \) Answer: The limit of the function \(g(x)\) at \(x = -1\) is -2.

Step 1: (Step 1: Analyzing the expression for x < -1)

Since the limit is when x approaches \(-1\), we have to analyze the first expression of the piecewise function which is \(\frac{x^2-1}{x+1}\) for \(x<-1\). To find the limit, we can try to simplify this expression and then substitute \(x=-1\).

Step 2: (Step 2: Simplify the expression)

We can factor the numerator of the expression \(\frac{x^2-1}{x+1}\) using the difference of squares formula. The difference of squares formula states that for any two numbers \(a\) and \(b\), \((a+b)(a-b)=a^2-b^2\). In this case, \(a = x\) and \(b = 1\). So, the expression can be simplified as: $$\frac{x^2-1}{x+1} = \frac{(x+1)(x-1)}{x+1}$$

Step 3: (Step 3: Cancel common factors)

After factoring the numerator, we can see that there is a common factor of \((x+1)\) both in the numerator and the denominator. We will cancel out this common factor, resulting in the simplified expression: $$\frac{(x+1)(x-1)}{x+1} = x-1$$

Step 4: (Step 4: Evaluate the limit)

Now that we have simplified the expression, we can substitute the limiting value \(x \rightarrow -1\) into the simplified expression \(x-1\): $$\lim_{x \rightarrow-1} g(x) =(-1) - 1=-2$$ Since the limit of the first expression as \(x\rightarrow-1\) is equal to the value of the second expression when \(x=-1\), we can conclude that the limit exists and is equal to: $$\lim_{x \rightarrow-1} g(x) = -2$$

Key Concepts

Piecewise Functions

Piecewise functions are a type of mathematical function that is defined by different expressions depending on the input value. The function from the exercise is an excellent example. It defines how the function behaves based on whether the input, denoted as \(x\), is less than or greater than or equal to \(-1\).
A piecewise function is usually written with different "pieces" or cases. Each piece specifies a function rule and a condition under which it should be applied. In this piecewise function, two different expressions dictate the value of \(g(x)\):

• The expression \(\frac{x^2-1}{x+1}\) is applicable for \(x < -1\).


• The expression is \(-2\) for \(x \geq -1\).

This structure allows piecewise functions to be versatile and model scenarios with abrupt changes, such as different rates or rules occurring at specific points. In the problem, the piecewise function directly impacts how we evaluate limits because we need to know which piece to use as \(x\) approaches \(-1\).
Understanding piecewise functions helps in analyzing how functions behave at specific points where the rule changes, like finding limits.

Difference of Squares

The difference of squares is a fundamental algebraic identity that simplifies expressions containing squared terms. It is given by \[(a+b)(a-b)=a^2-b^2.\]This identity is extremely helpful in factoring expressions, as seen in the exercise with \(x^2-1\). Factoring the difference of squares means recognizing that \(x^2-1\) can be rewritten as

• \((x+1)(x-1)\)

This is due to setting \(a = x\) and \(b = 1\). By doing so, we convert a polynomial expression into a format that might reveal common factors with the denominator. This step is crucial for simplifying expressions and finding limits, especially when expressions appear complex.
When dealing with limits, spotting differences of squares in the numerator can greatly aid in simplification by canceling terms, as was necessary here. Recognizing such patterns is critical for skillful algebraic manipulation.

Simplifying Expressions

Simplifying expressions is a key mathematical skill, particularly important when trying to find limits like \(\lim _{x \rightarrow-1} g(x)\). In order to simplify the expression \(\frac{x^2-1}{x+1}\), we used factoring, which transformed the expression into

• \((x+1)(x-1)/(x+1)\).

Simplification involves spotting and reducing common factors. Here, the \((x+1)\) in the numerator and the denominator can be canceled, leaving \(x-1\). This simplification significantly eases evaluating the limit as it reduces computation.
To evaluate the limit, substitute \(x\) with \(-1\) in the simplified expression, resulting in \(-2\). Simplifying expressions ensures our results are accurate and minimizes errors that come from complex algebraic manipulations. It's a toolkit essential for interpreting mathematical expressions, making them more comprehensible and manageable, especially around critical points like limits or asymptotes.