Problem 27 - Precalculus with Limits

Question

In Exercises 27 and 28, graph the function and find the limit(if it exists) as \(x\) approaches 2. \\[ f(x)= \left\\{ \begin{array}{rr} 2x + 1.& \mbox{if \(x < 2\)};\\\ x + 1.& \mbox{if \(x \geq 2\)}.\end{array} \right. \\]

Step-by-Step Solution

Answer

\(\lim_{{x \to 2}} f(x)\) does not exist.

Step 1: Graphing the function

First, divide the graphing process into two parts as per the conditions of the function. For \(x < 2\), the function is \(f(x) = 2x + 1\). For this region, use various \(x\) values less than 2 and find the corresponding \(f(x)\) values to plot the graph. For \(x \geq 2\), the function is \(f(x) = x + 1\). Similarly, for this region, take various \(x\) values greater than or equal to 2 and find the corresponding \(f(x)\) values to plot the graph.

Step 2: Evaluating the left-hand limit

The left-hand limit as \(x\) approaches 2, denoted as \(\lim_{{x \to 2^-}} f(x)\), refers to the limit of the function as \(x\) approaches 2 from the left side (values less than 2). Given that \(f(x) = 2x + 1\) for \(x < 2\), substitute 2 in \(2x + 1\), which gives 5. So, \(\lim_{{x \to 2^-}} f(x) = 5\)

Step 3: Evaluating the right-hand limit

The right-hand limit as \(x\) approaches 2, denoted as \(\lim_{{x \to 2^+}} f(x)\), refers to the limit of the function as \(x\) approaches 2 from the right side (values bigger than 2). Given that \(f(x) = x + 1\) for \(x \geq 2\), substitute 2 in \(x + 1\), which gives 3. So, \(\lim_{{x \to 2^+}} f(x) = 3\)

Step 4: Determining the limit of the function as \(x\) approaches 2

The limit of the function as \(x\) approaches 2, denoted as \(\lim_{{x \to 2}} f(x)\), exists only if the left-hand limit equals the right-hand limit. However, in this case, \(\lim_{{x \to 2^-}} f(x) = 5\) but \(\lim_{{x \to 2^+}} f(x) = 3\). As these limits are different, hence, \(\lim_{{x \to 2}} f(x)\) does not exist.

Key Concepts

Graphing Piecewise Functions

When it comes to graphing piecewise functions, visualizing the different segments is crucial. A piecewise function is defined by two or more expressions, each corresponding to a particular part of the domain. In our exercise, we have two expressions: for \(x < 2\), the function is defined as \(f(x) = 2x + 1\), and for \(x \thinspace\textgreater\thinspace\= 2\), it is \(f(x) = x + 1\).

To graph this function, start by plotting the first part for \(x < 2\). The line \(y = 2x + 1\) is straight, and its graph is a line with a slope of 2, crossing the y-axis at \(y = 1\). As \(x\) approaches 2, draw this line up to, but not including, the point where \(x = 2\). This often involves a small open circle to indicate that the function does not include the endpoint for this piece.

Next, graph the second part for \(x \thinspace\textgreater\thinspace\= 2\) by plotting \(y = x + 1\). This line has a slope of 1 and also crosses the y-axis at \(y = 1\), but here, the line includes the point where \(x = 2\), often marked with a filled-in dot to show inclusion. Where these two lines meet tells us about the limits and whether there is continuity at \(x = 2\).

Left-Hand Limit

The left-hand limit is a concept that is pivotal in understanding limits in precalculus. To evaluate the left-hand limit, we look at the value that the function approaches as \(x\) gets increasingly close to a certain point, in this case, 2, from the left side (values of \(x\) that are less than 2).

In our exercise, the left-hand limit as \(x\) approaches 2, written as \(\lim_{{x \to 2^-}} f(x)\), is determined by substuting \(x = 2\) into the part of the function that applies to values less than 2, which is \(f(x) = 2x + 1\). Doing this, we see that \(2(2) + 1 = 5\), and thus, the left-hand limit as \(x\) approaches 2 is 5. Conceptually, this means if you were to trace the graph from the left towards \(x = 2\), your pencil would point towards the \(y\)-value of 5.

Right-Hand Limit

Similarly, but from the other side, we have the right-hand limit. This is the value the function approaches as \(x\) gets increasingly close to a point from the right side (values of \(x\) that are greater than the point of interest).

In the given exercise, to find the right-hand limit as \(x\) approaches 2, we notate it as \(\lim_{{x \to 2^+}} f(x)\). Since we are considering values of \(x\) that are greater than or equal to 2, we look at the function \(f(x) = x + 1\). Plugging \(x = 2\) into this expression, we get \(x + 1 = 2 + 1 = 3\). So, the right-hand limit of the function as \(x\) approaches 2 is 3. If one imagines walking along the graph from the right towards \(x = 2\), their pencil would end pointing at the \(y\)-value of 3.

Evaluating Limits

Finally, when evaluating limits, it's about understanding what value, if any, a function is approaching as \(x\) moves towards a particular point. This involves considering both the left-hand and right-hand limits. If these two limits are equal, the limit at that point exists and is equal to this common value.

In this exercise, we've separately found that the left-hand limit as \(x\) approaches 2 is 5, while the right-hand limit is 3. Because these values are not the same, the overall limit of \(f(x)\) as \(x\) approaches 2 does not exist. This demonstrates a key principle of limits: for a limit to exist at a certain point, the left-hand and right-hand limits must agree. Since they don't in this case, we cannot say the function has a limit at \(x = 2\). This highlights a potential point of discontinuity or a 'jump' in the graph of the function.