Problem 6 - Calculus with Concepts in Calculus

Question

Find the given limit. $$ \lim _{t \rightarrow \infty} \frac{(t-1)(2 t+1)}{(3 t-2)(t+4)} $$

Step-by-Step Solution

Answer

The limit is \( \frac{2}{3} \).

Step 1: Understand the Limit and the Function

We are given the limit \( \lim _{t \rightarrow \infty} \frac{(t-1)(2 t+1)}{(3 t-2)(t+4)} \). Our task is to find the behavior of this function as \( t \) approaches infinity. We start by recognizing that the degree of the numerator and the denominator are the same.

Step 2: Expand the Polynomial Terms

First, expand both the numerator and denominator. The numerator is \((t-1)(2t+1) = 2t^2 + t - 2t - 1 = 2t^2 - t - 1\) and the denominator is \((3t-2)(t+4) = 3t^2 + 12t - 2t - 8 = 3t^2 + 10t - 8\). This gives us a clearer picture of the leading terms.

Step 3: Identify the Leading Terms

The leading term in the numerator \(2t^2 - t - 1\) is \(2t^2\). In the denominator \(3t^2 + 10t - 8\), the leading term is \(3t^2\). The limit is determined by these leading terms because as \(t\) approaches infinity, the lower degree terms become insignificant.

Step 4: Compute the Limit of Leading Terms

Focus on the highest power terms: \( \frac{2t^2}{3t^2} \). This simplifies the expression to \( \frac{2}{3} \). As \( t \to \infty \), this fraction approaches \( \frac{2}{3} \).

Step 5: Conclude the Limit Value

Since the other terms are negligible as \( t \) becomes very large and only the highest degree terms matter, the limit of the original expression is equivalent to the limit of these leading coefficients.

Key Concepts

Polynomial Expansion

When we're dealing with polynomials, sometimes their terms are neatly organized, but other times, they aren't. In such cases, we perform what's known as polynomial expansion. This process is essentially multiplying out expressions to combine like terms. In our given problem, we start with two polynomial expressions in the numerator \( (t-1)(2t+1) \) and in the denominator \( (3t-2)(t+4) \). By expanding these, we achieve a clearer understanding of each polynomial's terms.

• The expansion of \( (t-1)(2t+1) \) results in \( 2t^2 - t - 1 \).
• Meanwhile, \( (3t-2)(t+4) \) expands to \( 3t^2 + 10t - 8 \).

Recognizing how to expand polynomials is crucial because it allows you to identify which terms will be dominant, especially when analyzing behavior as \( t \) becomes very large.

Behavior at Infinity

Understanding the behavior of a polynomial function as its variable grows infinitely large, helps us understand the limit of a function at infinity. As \( t \) becomes very large in our original expression, only certain terms matter.As \( t \) approaches infinity, the lesser degree terms in polynomial expansions shrink in significance. This leaves us focusing primarily on the highest degree terms. It's like focusing on the tallest buildings in a city skyline—the other shorter buildings fade into the background.

• For the numerator \( 2t^2 - t - 1 \), the dominant term is \( 2t^2 \).
• For the denominator \( 3t^2 + 10t - 8 \), the dominant term is \( 3t^2 \).

Thus, the behavior of the function mimics the behavior of these leading terms, guiding us toward the function's limit as \( t \) becomes infinite.

Leading Terms in Polynomials

Leading terms in polynomials are those with the highest degree. These terms are the most significant when considering limits at infinity, because they grow the fastest as the variable increases.In our exercise, the leading terms significantly simplify the process of finding the limit:

• The leading term of the numerator, \( 2t^2 \), informs us that the dominant behavior is quadratic.
• Similarly, the leading term of the denominator, \( 3t^2 \), presents a quadratic behavior.

To find the limit at infinity, we look at the ratio of these leading terms:\[ \lim _{t \rightarrow \infty} \frac{2t^2}{3t^2} = \frac{2}{3} .\]By doing so, we've determined that as \( t \) grows, the function's limit approaches \( \frac{2}{3} \), thanks largely to the leading terms.