Problem 87 - Calculus: Early Transcendentals

Question

Finding a constant Suppose $$f(x)=\left\\{\begin{array}{ll} \frac{x^{2}-5 x+6}{x-3} & \text { if } x \neq 3 \\ a & \text { if } x=3 \end{array}\right.$$ Determine a value of the constant \(a\) for which \(\lim _{x \rightarrow 3} f(x)=f(3)\).

Step-by-Step Solution

Answer

Answer: The constant \(a\) should be equal to \(1\).

Step 1: Determine the limit of f(x) as x approaches 3

The function is given as a piecewise function with a different expression if \(x\neq 3\). Since we are finding the limit as \(x\) approaches \(3\), we will be focusing only on the expression when \(x\neq 3\), which is: $$\frac{x^{2}-5 x+6}{x-3}$$ To find the limit of this expression as \(x\) approaches \(3\), we can try to simplify it, possibly by factoring.

Step 2: Factor the expression

We can factor the numerator of the expression to simplify it: $$\frac{(x-2)(x-3)}{x-3}$$

Step 3: Simplify the expression

Now, since the factors \((x-3)\) appear in both the numerator and the denominator, we can cancel them out, leaving: $$\lim_{x\rightarrow 3} (x-2)$$

Step 4: Evaluate the limit

Now, we can find the limit by simply substituting the value of \(3\) for \(x\): $$\lim_{x\rightarrow 3} (x-2) = 3 - 2 = 1$$

Step 5: Solve for the constant a

Since we want the limit to be equal to \(f(3)\), we need to find the value of \(a\) such that: $$\lim_{x\rightarrow 3} f(x) = f(3)$$ We found in step 4 that \(\lim_{x\rightarrow 3} f(x) = 1\). Therefore, we need to solve for \(a\) in: $$1 = f(3)=a$$

Step 6: Find the value of a

The solution to the equation in the previous step is straightforward: $$a = 1$$ Thus, for the limit of the function \(f(x)\) as \(x\) approaches \(3\) to be equal to \(f(3)\), the constant \(a\) should be equal to \(1\).

Key Concepts

Piecewise Functions

Piecewise functions are mathematical expressions that have different formulas for different parts of their domain. They are like puzzles, where each piece fits into specific sections of the x-axis. In our exercise, the function \( f(x) \) provides one formula when \( x eq 3 \) and a different constant, \( a \), when \( x = 3 \). This setup is useful when a function behaves differently at a specific point compared to the rest of its domain.

When dealing with piecewise functions, especially in limits and continuity questions, it's crucial to inspect each piece's behavior and how they compare at transition points. In this problem, we are particularly interested in the behavior of \( f(x) \) as \( x \) approaches the transition point of 3. By simplifying and evaluating each part, piecewise functions help analyze situations where different rules apply under certain conditions.

• Always pay attention to the domain of each piece.
• Check for continuity at transition points.

Factorization

Factorization is a mathematical process of expressing a number or an algebraic expression as a product of its factors. In our exercise, the expression \( \frac{x^2 - 5x + 6}{x - 3} \) requires factorization of its numerator. By transforming the quadratic polynomial \( x^2 - 5x + 6 \) into \( (x-2)(x-3) \), we uncover its roots and simplify the expression.

This simplification step is crucial, especially when dealing with limits and indeterminate forms like \( \frac{0}{0} \). It allows us to cancel out factors and avoid undefined situations. Factorization not only aids in making a function easier to work with but also in understanding its underlying structure. When you see a polynomial that needs to be simplified, factorization is typically a key step:

• Identify common factors in expressions.
• Look for patterns like difference of squares or quadratic trinomials.

Continuity

Continuity is a fundamental concept in calculus that ensures a function behaves in a predictable manner as you approach a specific point. A function is continuous at a point if three criteria are met:

• The function is defined at that point.
• The limit of the function as it approaches the point exists.
• The limit value equals the function's value at that point.

In our piecewise function problem, we apply these criteria by ensuring the limit \( \lim_{x \rightarrow 3} f(x) \) equals \( f(3) \). By setting the value of \( a \) equal to the limit, we ensure \( f(x) \) is continuous at \( x = 3 \).

Continuity allows us to bridge gaps between different parts of a function, ensuring no sudden jumps or breaks at specified points. In practical terms, a continuous function provides a smooth curve or line on a graph, which is essential for accurate interpretations and predictions.