Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the limits. \(\lim _{y \rightarrow-\infty} \frac{9 y^{3}+1}{y^{2}-2 y+2} .\) Hint: Divide numerator and denominator by \(y^{2}\).

Step-by-Step Solution

Answer

The limit is negative infinity.

Step 1: Identify Highest Powers

To find the limit as \( y \) approaches negative infinity, we first identify the highest powers of \( y \) in both the numerator and the denominator. In the numerator, the highest power is \( y^3 \), and in the denominator, the highest power is \( y^2 \).

Step 2: Rewrite Expression by Dividing

According to the hint, divide each term in the numerator and the denominator by \( y^2 \). This gives \[ \frac{9y^3 + 1}{y^2 - 2y + 2} = \frac{9 \frac{y^3}{y^2} + \frac{1}{y^2}}{\frac{y^2}{y^2} - 2 \frac{y}{y^2} + \frac{2}{y^2}}. \] Simplifying the fractions yields \[ \frac{9y + \frac{1}{y^2}}{1 - \frac{2}{y} + \frac{2}{y^2}}. \]

Step 3: Evaluate the Limit

As \( y \to -\infty \), the terms \( \frac{1}{y} \) and \( \frac{1}{y^2} \) approach 0. This simplifies the expression to \[ \lim_{y \to -\infty} \frac{9y + 0}{1 + 0 + 0} = \lim_{y \to -\infty} 9y. \]

Step 4: Determine Result

Since \( 9y \to -\infty \) as \( y \to -\infty \), the limit evaluates to negative infinity.

Key Concepts

Polynomial Functions

Polynomial functions are mathematical expressions that involve sums of terms, each consisting of a variable raised to a whole number power, along with a coefficient. For example, the expression \(9y^3 + 1\) is a polynomial function. It has two terms: \(9y^3\) and \(1\). The highest power of the variable, \(y\), dictates the degree of the polynomial. Here, the highest power is 3, so this is a degree 3 polynomial.
Polynomials are very flexible and can model a wide range of behaviors. They can have peaks, valleys, and can stretch out toward infinity. They are continuous and smooth, which means there are no sharp corners or holes. This continuity makes them easier to work with in calculus when calculating limits, derivatives, or integrals.

When dealing with limits in polynomial functions as a variable approaches infinity or negative infinity, the leading term (the term with the highest power) dominates the behavior of the polynomial. This is because as the variable becomes very large or very small, the highest power's effect outpaces that of all lower power terms. So, a polynomial like \(9y^3 + 1\) behaves similarly to \(9y^3\) when \(y\) is very large or very small in absolute value.

Rational Functions

Rational functions are ratios (fractions) of two polynomials. In our case, we have \(\frac{9y^3 + 1}{y^2 - 2y + 2}\) as a rational function. The numerator is a polynomial, \(9y^3 + 1\), and the denominator is another polynomial, \(y^2 - 2y + 2\).
Rational functions can exhibit various behaviors based on the degrees of the numerator and the denominator. These functions can have horizontal asymptotes, vertical asymptotes, and even oblique asymptotes. The behavior of a rational function as the variable approaches infinity can be particularly interesting.

When finding a limit as you approach infinity, often it helps to consider the leading terms of both the numerator and the denominator. The degree of these terms will influence the limit greatly. If the numerator has a higher degree than the denominator, as we see here with \(9y^3\) versus \(y^2\), the function will tend toward positive or negative infinity, depending on the sign of the leading coefficient.

Infinite Limits

Infinite limits occur when the function's values increase or decrease without bound as the variable approaches a certain point or infinity. In the limit \(\lim_{y \to -\infty} \frac{9y^3 + 1}{y^2 - 2y + 2}\), we examine what happens to the function as \(y\) becomes very large in the negative direction.
In order to simplify and find the limit, we divide every term by \(y^2\), the highest power in the denominator. This step simplifies the equation and isolates the dominating behavior of the function.

After simplification, we're left with \(\frac{9y + 0}{1 - \frac{2}{y} + \frac{2}{y^2}}\). As \(y\) approaches negative infinity, the fractions \(\frac{1}{y}\) and \(\frac{1}{y^2}\) tend towards zero, simplifying our expression to just \(9y\). Since \(9y\) becomes indefinitely large negatively, the limit and thus the infinite limit, is \(-\infty\). This direct relationship between the leading term and the behavior at infinity is a core concept in understanding infinite limits in calculus.