Problem 42
Question
Find the limit, if it exists. If the limit does not exist, explain why. \( \displaystyle \lim_{x \to -6}\frac{2x + 12}{| x + 6 |} \)
Step-by-Step Solution
Verified Answer
The limit does not exist because the left-hand and right-hand limits differ.
1Step 1: Identify the critical point
First, we identify the critical point where the limit is evaluated: \( x = -6 \). At this point, the expression \(|x + 6|\) tends to zero. This indicates that the expression might be undefined unless we consider separate approaches as \(x\) approaches from different sides (left and right).
2Step 2: Evaluate the left-hand limit
Evaluate the limit as \(x\) approaches \(-6\) from the left (denoted \(x \to -6^-\)). Here, \(x < -6\) and therefore \(|x + 6| = -(x + 6)\). Thus, the function becomes:\[ \frac{2x + 12}{-(x + 6)} = \frac{2(x + 6)}{-(x + 6)} = -2 \]The left-hand limit is \(-2\).
3Step 3: Evaluate the right-hand limit
Evaluate the limit as \(x\) approaches \(-6\) from the right (denoted \(x \to -6^+\)). In this case, \(x > -6\) which implies \(|x + 6| = x + 6\). Thus, the function simplifies to:\[ \frac{2x + 12}{x + 6} = \frac{2(x + 6)}{x + 6} = 2 \]The right-hand limit is \(2\).
4Step 4: Compare the left-hand and right-hand limits
Compare the results from Steps 2 and 3. The left-hand limit as \(x\) approaches \(-6\) is \(-2\), while the right-hand limit is \(2\). Since these two limits are not equal, the overall limit does not exist.
Key Concepts
Left-Hand LimitRight-Hand LimitPiecewise Functions
Left-Hand Limit
The left-hand limit is all about understanding what happens to a function as it approaches a certain point from the left side. Here, we're talking about approaching from values that are smaller than the target point. Let's say you have a function, and you want to find the limit as the input variable approaches some critical value, say \(x = c\).
To find the left-hand limit, you would use \(x \to c^-\). This notation means that the function is being evaluated as \(x\) comes very close to \(c\) but from the left.
To find the left-hand limit, you would use \(x \to c^-\). This notation means that the function is being evaluated as \(x\) comes very close to \(c\) but from the left.
- The key feature is that these values are slightly less than \(c\).
- For piecewise functions, this might involve a specific expression that applies only when approaching from the left.
Right-Hand Limit
The right-hand limit works similar to the left-hand limit but from the other side. With the right-hand limit, we look at what happens as the input variable of a function approaches a certain point from the right side, or from values that are slightly larger than the target point.
Here, the notation used is \(x \to c^+ \) which indicates approaching the critical value \(x = c \) from the right.
Here, the notation used is \(x \to c^+ \) which indicates approaching the critical value \(x = c \) from the right.
- This involves considering values just a bit greater than \(c\).
- In piecewise functions, there can be a separate rule or formula applied when approaching from the right.
Piecewise Functions
Piecewise functions are a fascinating part of mathematics and involve functions defined by multiple sub-functions, each with its own specific formula. These pieces are defined over various intervals of the input variable, like cutting a pizza into slices, where each slice has a different topping.
When working with piecewise functions, one of the critical aspects to consider is how each segment behaves around the points where the formula changes. This is where limits, especially left-hand and right-hand limits, come into play.
When working with piecewise functions, one of the critical aspects to consider is how each segment behaves around the points where the formula changes. This is where limits, especially left-hand and right-hand limits, come into play.
- Each section of a piecewise function may have different slopes, curvatures, and intercepts.
- Limit evaluations help determine continuity or discontinuity at the joining points.
Other exercises in this chapter
Problem 42
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