Problem 59 - Calculus and its applications

Question

Is the function given by $$ G(x)=\left\\{\begin{array}{ll} \frac{x^{2}-4}{x-2}, & \text { for } x \neq 2, \\ 5, & \text { for } x=2 \end{array}\right. $$ continuous at \(x=2 ?\) Why or why not?

Step-by-Step Solution

Answer

No, the function is not continuous at \(x = 2\) because \(G(2)\) does not equal \(\lim_{{x \to 2}} G(x)\).

Step 1: Determine the value of G at x=2

Directly substitute the value into the piecewise function. For \(x = 2\), the function \(G(x)\) is given explicitly as 5. So, \(G(2) = 5\).

Step 2: Setup the limit for x approaching 2

To find if \(G(x)\) is continuous at \(x = 2\), we must determine \(\lim_{{x \to 2}} G(x)\). For \(x eq 2\), \(G(x) = \frac{x^{2}-4}{x-2}\). This expression must be simplified to evaluate the limit.

Step 3: Simplify the expression

The expression \(\frac{x^{2}-4}{x-2}\) can be simplified. Notice that \(x^{2} - 4\) can be factored as \((x + 2)(x - 2)\), so the expression becomes:\[\frac{(x + 2)(x - 2)}{x - 2} = x + 2, \text{ for } x eq 2\].

Step 4: Evaluate the limit

Now we compute the limit \(\lim_{{x \to 2}} (x + 2)\). Substituting \(x = 2\) into \(x + 2\) gives \(2 + 2 = 4\). Therefore, the limit as \(x\) approaches 2 is 4.

Step 5: Compare the limit and the function value

For continuity at \(x = 2\), \(G(2)\) must equal \(\lim_{{x \to 2}} G(x)\). We found \(G(2) = 5\) and \(\lim_{{x \to 2}} G(x) = 4\). Since these values are not equal, \(G(x)\) is not continuous at \(x = 2\).

Key Concepts

Piecewise Functions

Piecewise functions are functions defined by different expressions depending on the input value. This means they are defined in pieces. A common everyday analogy is a train schedule. Trains run at specified times and place stops at different intervals, so their schedule can be thought of in pieces too.
In mathematics, piecewise functions are often encountered when modeling situations that have conditions or changes at specific points. These changes could be real-world phenomena like tax rates, tariffs, or even speed regulations on a busy street.

For analyzing a piecewise function, each piece or segment is defined over an interval of the domain. When working with these functions, it's crucial to pay attention to endpoints. You must plug in values directly as per the function definition for segments, especially at points where the function definition switches. In our original problem, the function is defined differently when \(x = 2\) and \(x eq 2\). This enables a careful tailoring of the function's behavior around specific points.

Limits in Calculus

Limits in calculus explore the behavior of functions as they get close to certain points. They give us an idea of the value that a function approaches as the input approaches a particular number. To understand if a piecewise function is continuous at a given point, one needs to calculate the limit at that point.

• The concept of a limit is foundational for defining derivatives and integrals, core concepts within calculus.
• Evaluating limits help ascertain smooth transitions (or not) in piecewise functions.

To determine if our function \(G(x)\) is continuous at \(x=2\), we calculated \(\lim_{{x \to 2}} G(x)\). Since the result did not match the defined function value at that point, we concluded that the function was not continuous. Thus, limits provide a tool to check the consistency and predictability of function behavior in calculus.

Factoring Expressions

Factoring is a process used in algebra to simplify expressions, making computations more straightforward and insights into the expression's structure clearer. It involves breaking down a complicated expression into simpler 'factors' that, when multiplied together, produce the initial expression.
In the problem, an essential step was factoring the expression \(x^2 - 4\). This expression can be recognized as a difference of squares, a common type of expression in algebra; it can be rewritten as \((x + 2)(x - 2)\) because both terms involve a square.

When finding limits or roots, factoring helps to cancel out terms that may cause indeterminate forms like division by zero. After factoring, the troublesome denominator \(x - 2\) in \(\frac{(x + 2)(x - 2)}{x - 2}\) was simplified to just \(x + 2\), since \(x eq 2\). This allowed an easy evaluation of the limit at \(x = 2\). Understanding factoring is invaluable in algebra and calculus, as it opens the door to simplifying and resolving complex expressions.