Problem 24 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the indicated one-sided limit, if it exists. \(\lim _{x \rightarrow 1^{+}} \frac{x+2}{x+1}\)

Step-by-Step Solution

Answer

The one-sided limit (right-hand limit) of the given function as x approaches 1 is \(\frac{3}{2}\).

Step 1: Understand the given function and the limit

The function is given as \(\frac{x+2}{x+1}\) and we need to find the one-sided limit as \(x \rightarrow 1^+\). It means we need to find the value of the function when x approaches 1 from the positive side.

Step 2: Substitute the value of x and check if the function is continuous

Substitute \(x = 1\) in the given function to see if it's continuous and doesn't have any undefined values at the specified point. Function: \(\frac{x+2}{x+1}\) Substituting \(x = 1\): \[\frac{1 + 2}{1 + 1} = \frac{3}{2}\] The function is continuous at \(x = 1\) since the denominator is not zero. It means the limit exists.

Step 3: Calculate the one-sided limit

Since the function is continuous at \(x = 1\), the right-hand limit is equal to the value of the function when \(x = 1\). Thus, \(\lim_{x \rightarrow 1^+} \frac{x+2}{x+1} = \frac{3}{2}\). The one-sided limit (right-hand limit) of the given function as x approaches 1 is \(\frac{3}{2}\).

Key Concepts

One-sided limits

In calculus, one-sided limits are an important tool to understand how functions behave as they approach a specific point from one direction. When we talk about a one-sided limit, we're referring to either a right-hand limit or a left-hand limit.

• The right-hand limit, denoted as \( \lim_{x \to a^+} f(x) \), describes the value that the function \( f(x) \) approaches as \( x \) approaches \( a \) from values greater than \( a \).
• On the other hand, the left-hand limit, displayed as \( \lim_{x \to a^-} f(x) \), identifies the value the function approaches as \( x \) moves toward \( a \) from values less than \( a \).

Understanding one-sided limits is crucial because sometimes, the behavior of a function differs depending on the direction from which \( x \) approaches the point. For instance, in the exercise provided, \( \lim_{x \to 1^+} \frac{x+2}{x+1} \) means we examine \( \frac{x+2}{x+1} \) as \( x \) approaches 1 from the positive, or right-hand, side. Grasping one-sided limits deepens our understanding of a function's behavior at points of discontinuity or sharp turns.

Continuity

Continuity in a function is an essential aspect to determine the behavior of functions around certain points. A function is said to be continuous at a point if three conditions are satisfied:

• The function \( f(x) \) is defined at the point \( x = a \).
• The limit \( \lim_{x \to a} f(x) \) exists.
• The limit of the function as \( x \) approaches \( a \) equals the function's value at \( a \), i.e., \( \lim_{x \to a} f(x) = f(a) \).

When a function is continuous at a point, it means there are no gaps, jumps, or holes at that point, making it easier to calculate limits. In the provided exercise, when we substitute \( x = 1 \) into the function \( \frac{x+2}{x+1} \), the result \( \frac{3}{2} \) confirms there is no discontinuity at \( x = 1 \), as the function is well-defined and the denominator is not zero. This continuity means the one-sided limit from the right is straightforward to calculate and matches the function's value at \( x = 1 \).

Limit calculation

The process of limit calculation allows us to predict a function's behavior near a particular point, even if a direct computation at that point might be tricky. For straightforward functions, if a direct substitution of the approaching value into the function doesn’t result in division by zero or indeterminate forms, the limit can be calculated directly.Here's how you generally calculate limits:

• If substituting the point into the function gives a real number without issues, then the limit equals this value.
• If you encounter an indeterminate form like \( \frac{0}{0} \), you might need to simplify the expression or use techniques like L'Hôpital's Rule, factoring, or trigonometric identities, depending on the problem.

In the given problem, substituting \( x = 1 \) directly into the function \( \frac{x+2}{x+1} \) yields \( \frac{3}{2} \), indicating that there were no complexities or complications at this point. Therefore, the one-sided limit as \( x \to 1^+ \) is quickly determined to be \( \frac{3}{2} \). This seamless calculation underscores the benefit of dealing with continuous functions. Limit calculations thus provide precise ways to establish function behavior close to points of interest.