Problem 49 - Calculus: Early Transcendentals

Question

Let \( g(x) = \frac{x^2 + x - 6}{| x - 2 |} \). (a) Find (i) \( \displaystyle \lim_{x \to 2^+}g(x) \) (ii) \( \displaystyle \lim_{x \to 2^-}g(x) \) (b) Does \( \displaystyle \lim_{x \to 2}g(x) \) exist? (c) Sketch the graph of \( g \).

Step-by-Step Solution

Answer

(a) (i) 5, (ii) -5. (b) No, the limit does not exist.

Step 1: Analyze the function

The function is given as \( g(x) = \frac{x^2 + x - 6}{| x - 2 |} \). The expression \( x^2 + x - 6 \) can be factored into \((x - 2)(x + 3)\). So, the function becomes \( g(x) = \frac{(x-2)(x+3)}{|x-2|} \). This helps us understand how the numerator simplifies and the role of the absolute value in the denominator.

Step 2: Simplify for right-hand limit

For \( x \to 2^+ \), the absolute value \(|x - 2| = x - 2 \), because \(x - 2 \) is positive for \( x > 2 \). Therefore, the function simplifies to \( g(x) = x + 3 \) as \( x \to 2^+ \).

Step 3: Evaluate the right-hand limit

Now, we evaluate the limit \( \lim_{x \to 2^+}(x + 3) \). By direct substitution, we find \( \lim_{x \to 2^+}(x + 3) = 2 + 3 = 5 \).

Step 4: Simplify for left-hand limit

For \( x \to 2^- \), the absolute value \(|x - 2| = -(x - 2) \), because \( x - 2 \) is negative for \( x < 2 \). Therefore, the function simplifies to \( g(x) = -(x + 3) \) as \( x \to 2^- \).

Step 5: Evaluate the left-hand limit

Now, we evaluate \( \lim_{x \to 2^-}-(x + 3) \). By direct substitution, we find \( \lim_{x \to 2^-}-(x + 3) = -(2 + 3) = -5 \).

Step 6: Compare one-sided limits

We have \( \lim_{x \to 2^+}g(x) = 5 \) and \( \lim_{x \to 2^-}g(x) = -5 \). The left-hand limit and the right-hand limit are different.

Step 7: Determine existence of overall limit

Since \( \lim_{x \to 2^+} e \lim_{x \to 2^-} \), the overall limit \( \lim_{x \to 2}g(x) \) does not exist.

Step 8: Sketch the graph of the function

To sketch the graph, note that for \( x > 2 \), the function behaves as \( y = x + 3 \), a straight line with a slope of 1 passing through \((2, 5)\). For \( x < 2 \), it behaves as \( y = -(x + 3) \), a line with a slope of -1 crossing the y-axis at \(-3\). There is a discontinuity (jump) at \( x = 2 \), with no well-defined value at this point.

Key Concepts

One-Sided Limits

Understanding one-sided limits is crucial when dealing with functions where the behavior changes around a specific point. In our exercise, we analyze the limits of the function \( g(x) = \frac{x^2 + x - 6}{|x - 2|} \) as \( x \) approaches 2 from the right (denoted \( x \to 2^+ \)) and from the left (denoted \( x \to 2^- \)).

• For \( x \to 2^+ \), we note that \(|x - 2|\) equals \(x - 2\), causing the function to simplify to \(g(x) = x + 3\) at points just above 2.
• For \( x \to 2^- \), \(|x - 2|\) equals \(-(x - 2)\), resulting in \(g(x) = -(x + 3)\) at points just below 2.

In essence, the behavior of the function is influenced by the direction from which \( x \) approaches the point of interest. Calculating these limits yields values of 5 and -5 respectively, confirming different behaviors on each side.

Discontinuous Functions

A function is discontinuous at a point if the left-hand limit and right-hand limit as \( x \) approaches that point are not equal. In our case, \( g(x) \) is discontinuous at \( x = 2 \). The right-hand limit \( \lim_{x \to 2^+}g(x) = 5 \), and the left-hand limit \( \lim_{x \to 2^-}g(x) = -5 \). These differing values mean that there is a "jump" in the function at \( x = 2 \) since the function does not converge to a single value. Therefore, \( \lim_{x \to 2}g(x) \) does not exist as the complete limits from either side do not meet. Understanding discontinuities is important in calculus because they signify points where a function cannot be easily defined, requiring additional analysis to describe accurately.

Graph Sketching

Sketching graphs of functions involves identifying key characteristics such as intercepts, slopes, discontinuities, and behavior as \( x \) approaches certain values. For \( g(x) \), the task involves sketching two linear functions with clear guidelines:

• For \( x > 2 \), sketch \( y = x + 3 \). This line has a positive slope and a y-intercept at 3, passing through the point (2, 5).
• For \( x < 2 \), sketch \( y = -(x + 3) \). This line has a negative slope and crosses the y-axis at -3.

The graphs of these individually might appear straightforward, but it's important to correctly convey the change in behavior at \( x = 2 \) without connecting these lines, showing clearly the jump where no value at \( x = 2 \) can be consistently defined.

Piecewise Functions

The function \( g(x) = \frac{x^2 + x - 6}{|x - 2|} \) can be thought of as a piecewise function. Piecewise functions are defined by different expressions depending on the domain of input values. This type of function is seen in practice when a function's behavior is controlled by conditions, often leading to discontinuities. For example, the function is:

• \( g(x) = x + 3 \) when \( x > 2 \)
• \( g(x) = -(x + 3) \) when \( x < 2 \)

By treating \( g(x) \) as piecewise, it becomes easier to see how the calculations for one-sided limits were derived and why the behavior of \( g(x) \) changes so drastically at \( x = 2 \). Recognizing functions as piecewise can often simplify complex calculus problems by letting you focus on distinct parts or pieces of the overall function and how they interact at certain points.