Problem 73 - Calculus: Early Transcendentals

Question

One-sided limits Let $$f(x)=\left\\{\begin{array}{ll} x^{2}+1 & \text { if } x<-1 \\ \sqrt{x+1} & \text { if } x \geq-1 \end{array}\right.$$ Compute the following limits or state that they do not exist. a. \(\lim _{x \rightarrow-1^{-}} f(x)\) b. \(\lim _{x \rightarrow-1^{+}} f(x) \quad\) c. \(\lim _{x \rightarrow-1} f(x)\)

Step-by-Step Solution

Answer

Explain your answer based on the one-sided limits. Answer: No, the limit of the function \(f(x)\) does not exist at \(x=-1\). This is because the left-side limit at \(x=-1\) is 2, while the right-side limit is 0. Since the left-side and right-side limits are not equal, the limit of \(f(x)\) at \(x=-1\) does not exist.

Step 1: Identify the appropriate function definition for the left side limit

As we are considering the left side limit (\(x \rightarrow -1^{-}\)), we should use the definition which is valid for \(x < -1\). In this case, that definition is \(f(x) = x^2 + 1\).

Step 2: Calculate the limit using the definition

Now we will substitute \(x\) with the limit value of \(-1\) and calculate the result: $$\lim_{x \rightarrow -1^-} f(x) = (-1)^2 + 1 = 1 + 1 = 2$$ The left-side limit at \(x=-1\) is 2. #b. Finding the right-side limit at x = -1#

Step 1: Identify the appropriate function definition for the right side limit

As we are considering the right side limit (\(x \rightarrow -1^{+}\)), we should use the definition which is valid for \(x \geq -1\). In this case, that definition is \(f(x) = \sqrt{x+1}\).

Step 2: Calculate the limit using the definition

Now we will substitute \(x\) with the limit value of \(-1\) and calculate the result: $$\lim_{x \rightarrow -1^+} f(x) = \sqrt{(-1)+1} = \sqrt{0} = 0$$ The right-side limit at \(x=-1\) is 0. #c. Checking if the limit exists at x = -1#

Step 1: Compare the left and right side limits

In order to determine if the limit exists at \(x = -1\), both the left-side and right-side limits must be equal. In this case, $$\lim_{x \rightarrow -1^-} f(x) = 2$$ and $$\lim_{x \rightarrow -1^+} f(x) = 0$$ are not equal.

Step 2: State the conclusion

Since the left-side limit and the right-side limit at \(x=-1\) are not the same, the limit \(\lim_{x \rightarrow -1} f(x)\) does not exist.

Key Concepts

Understanding Left-Sided Limits

A left-sided limit is concerned with the behavior of a function as the input approaches a specific point from the left side. Specifically, this involves analyzing what happens as we approach the point with values less than that point. Let's break it down further with our exercise.

• For the function \(f(x)\), when \(x\) approaches \(-1\) from the left, we look at values less than \(-1\).
• According to the piecewise function, the applicable formula is \(f(x) = x^2 + 1\).
• Thus, the left-sided limit is calculated by substituting \(-1\) into \(x^2 + 1\), giving us 2 as the result.

It's important to understand that the left-sided limit shows the function's trend as it gets closer to the point from the left, but not at the point itself. This helps in determining continuity and behavior of functions in mathematical analysis.

Exploring the Right-Sided Limit

The right-sided limit examines how a function behaves as its input approaches a particular value from the right. This means considering values that are equal to or greater than this point. Let's explore this concept with the given problem.

• In our example, as \(x\) approaches \(-1\) from the right, we deal with values that are greater than or equal to \(-1\).
• For these values, the function is defined by the equation \(f(x) = \sqrt{x+1}\).
• By substituting \(-1\) into \(\sqrt{x+1}\), we find that the right-sided limit is 0.

Understanding the right-sided limit is crucial since it gives insight into what the function approaches as the variable increases to the target point. Recognizing differences between left and right-sided limits helps in analyzing whether a full limit exists or not.

What is a Piecewise Function?

A piecewise function is one of the versatile tools in mathematics, allowing flexibility and specificity in describing functions that have distinct definitions for different intervals of their domain.

• In our exercise, the function \(f(x)\) has different rules for \(x < -1\) and \(x \geq -1\).
• This on-off behavior is captured by the various pieces, each valid over a defined range.
• Such functions are useful when a phenomenon exhibits different behaviors under different conditions.

Piecewise functions are common in real life, modeling scenarios where conditions change—like pricing strategies, physical systems, or even daylight hours. Understanding how each piece operates helps prevent mistakes and allows a more precise analysis of limits and behaviors at junction points like \(x = -1\). In calculus, this understanding paves the way for further studies in continuity and differentiability, crucial in evaluating complex functions.