Problem 10 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

\(f(x)=\frac{(x+2)\left(x^{2}-x\right)}{x(x-1)}\) (a) \(\lim _{x \rightarrow 0} f(x)\) (b) \(\lim _{x \rightarrow 1} f(x)\)

Step-by-Step Solution

Answer

The limit of \(f(x) = (x+2)\) as \(x \rightarrow 0\) is 2 and as \(x \rightarrow 1\) is 3.

Step 1: Simplify the Expression

We commence by simplifying the function \(f(x)=\frac{(x+2)(x^{2}-x)}{x(x-1)}\). We notice that the terms \(x^{2}-x\) in the numerator can simplify with the terms \(x(x-1)\) in the denominator. Simplifying gives us: \(f(x) = (x+2).\)

Step 2: Finding Limit at 0

Now we will find \(\lim _{x \rightarrow 0} f(x)\). Substitute x = 0 into \(f(x) = (x+2)\), the simplified function obtained in previous step. We get the result: \(\lim _{x \rightarrow 0} f(x) = 2.\)

Step 3: Finding Limit at 1

Similarly, we find \(\lim _{x \rightarrow 1} f(x)\) by substituting x = 1 into the simplified function \(f(x) = (x+2)\). The result we get is: \(\lim _{x \rightarrow 1} f(x) = 3.\)

Key Concepts

Limits and Continuity

When we talk about limits in calculus, we're discussing what happens to a function as the input approaches a certain value. It's a foundational concept that lets us analyze a function's behavior without necessarily reaching that point. Particularly, the notion of continuity is tied up with limits — a function is continuous at a point if the limit as the input approaches that value matches the function's value at that point.

To ensure continuity, a function must meet three criteria: it must be defined at the point in question; the limit as x approaches the value must exist; and the limit must equal the function's value at that point. Our exercise demonstrates continuity at x=0 after simplification since the limit equals the function's value, which is 2.

Simplifying Rational Expressions

Rational expressions, such as the ones in our exercise, can often be simplified by factoring and reducing common terms. Simplifying these expressions helps with evaluating limits and analyzing function behavior. In the provided exercise, we observed that the terms in the numerator and denominator could be factored and common terms could be canceled out to simplify the function.

This simplification step is not just for cleanliness; it's crucial for being able to find the limit, especially when direct substitution would lead to an undefined term, like division by zero. In our case, simplification led to an uncomplicated linear function, which significantly eased the process of finding the limits.

Substitution in Limits

Substitution is usually the first strategy tried when evaluating limits. If a function is continuous at the point to which x is approaching, you can substitute the value directly into the function to find the limit.

This strategy was applied in our exercise after simplifying the rational expression to \(f(x) = (x+2)\), which is continuous everywhere. For \(x=0\), substituting 0 into \(f(x)\) directly gave us the limit. Similarly, for \(x=1\), substitution provided the limit immediately. This step is often straightforward but relies on the function being continuous at the point in question.

Limit Evaluation Strategies

Evaluating limits might involve several strategies, especially when direct substitution isn't possible. Besides substituting, we might factoring, rationalizing, or applying L'Hopital’s Rule when faced with indeterminate forms like 0/0.

Our exercise didn't require these advanced tactics due to the simplicity achieved through factoring. Nonetheless, it's crucial to note that limits can be approached from both the left (\(x\) approaching from values less than the point) and the right (\(x\) approaching from values greater than the point), and the limit exists only if both sides agree. The ease of limit calculation in this exercise shouldn't overshadow the need to understand and deploy a variety of strategies for different cases of limit problems.