Question

Determine whether the function is continuous on the entire real line. Explain your reasoning. \(g(x)=\frac{x^{2}-4 x+4}{x^{2}-4}\)

Short answer

The function \(g(x)=\frac{x^{2}-4 x+4}{x^{2}-4}\) is continuous everywhere on its domain except at \(x = -2\).

Step-by-step solution

Identify the problematic points

Set the denominator equal to zero and solve for \(x\):\(x^{2} - 4 = 0\)\((x-2)(x+2) = 0\)So, \(x = 2\) or \(x = -2\) are the points where the function could be discontinuous.

Check the limit at \(x = 2\)

Take the limit of the function as \(x\) approaches 2:\(\lim_{{x \to 2}} \frac{x^{2}-4 x+4}{x^{2}-4}= \lim_{{x \to 2}} \frac{(x-2)^{2}}{(x-2)(x+2)}\)After the cancellation of \((x-2)\), we have:\(\lim_{{x \to 2}} \frac{x-2}{x+2}=0\)which is a real number. So, the function can be made continuous at \(x = 2\) by redefining it to be 0 at that point.

Check the limit at \(x = -2\)

Similarly, take the limit of the function as \(x\) approaches -2:\(\lim_{{x \to -2}} \frac{x^{2}-4 x+4}{x^{2}-4}= \lim_{{x \to -2}} \frac{(x-2)^{2}}{(x-2)(x+2)}\)After the cancellation of \((x+2)\), we have:\(\lim_{{x \to -2}} \frac{x-2}{-2x}=\infty\)which is not a real number. So, the function cannot be made continuous at \(x = -2\).

Key concepts

Limits

Limits are fundamental in understanding the behavior of functions as they approach specific points. In mathematics, when we say a function has a limit as it nears a specific point, it implies that the values of the function can get arbitrarily close to some number as the input gets closer to a particular point. This concept helps us explore the behavior of the function around points where the function may not be explicitly defined.
Imagine you're approaching a stoplight. The limit helps us understand what would happen as you get closer, even if you never actually stop right at the light.
For the function in our exercise, we examine the limits as the input, \(x\), approaches points where the function might not be defined—here, \(x = 2\) and \(x = -2\).

• At \(x = 2\), the limit is zero, suggesting the function can be nicely smoothed out by redefining it.
• At \(x = -2\), however, the limit is not defined, indicating a breakdown or an infinite behavior which limits our ability to make such adjustments.

Understanding limits gives us valuable insight into potential discontinuities and function behavior around these points.

Rational functions

Rational functions are a special category of functions defined by the ratio of two polynomials. These functions can exhibit complex behaviour, especially where their denominator equals zero, often leading to potential points of discontinuity.
To better understand, think of rational functions as a fraction \( \frac{P(x)}{Q(x)} \), where both \(P(x)\) and \(Q(x)\) are polynomials. The function is problematic if and when \(Q(x) = 0\), since division by zero is undefined in mathematics.
Our exercise function, \(g(x)=\frac{x^{2}-4 x+4}{x^{2}-4}\), illustrates this brilliantly. It combines understanding both the numerator and the denominator's behavior. When the denominator equals zero, specifically at \(x = 2\) and \(x = -2\), interesting things happen.

• By factoring, we see that the numerator and denominator share a common term, which allows us to simplify and examine the behavior of \(g(x)\) at these troublesome points.
• This simplification plays a crucial role in determining continuous behavior or discontinuity across the real line.

Recognizing and working with rational functions is key to deepening your understanding of function properties, including continuity and discontinuity.

Discontinuity

Discontinuity in functions occurs where a function cannot be continued or smoothed out seamlessly. In simpler terms, it is where the function "jumps," or blows up, or simply where there is a hole in its graph.
Analyzing the function \(g(x)=\frac{x^{2}-4 x+4}{x^{2}-4}\), reveals discontinuity at points where the denominator vanishes, leading to undefined behavior.

• At \(x = 2\), the discontinuity arises from a removable gap—a hole can be "filled" by redefining the function, since the limit exists.
• Conversely, at \(x = -2\), the discontinuity presents as a vertical asymptote, indicating that the function cannot be adjusted to be continuous. The behavior here is infinite, leading to a breakdown that cannot be addressed by simple fixes.

Understanding the nature of discontinuity helps in assessing whether a function can be made continuous by redefining or adjusting its expression. It is essential for gauging the complete picture of how a function behaves across its domain.