Question

Evaluate the given limit. $$ \lim _{x \rightarrow-\infty} \frac{x^{3}+2 x^{2}+1}{5-x^{2}} $$

Short answer

The limit is \( \infty \).

Step-by-step solution

Identify the dominant terms

When evaluating limits at infinity, we compare the highest degree polynomial terms in the numerator and the denominator because they have the greatest impact as \( x \) approaches infinity. In the expression \( \frac{x^3 + 2x^2 + 1}{5 - x^2} \), the dominant term in the numerator is \( x^3 \), and in the denominator, it is \( -x^2 \).

Simplify the expression by dividing by the highest power of x in the denominator

Divide each term of the numerator and denominator by \( x^2 \), the highest power of \( x \) in the denominator: \[\lim_{x \to -\infty} \frac{x^3/x^2 + 2x^2/x^2 + 1/x^2}{5/x^2 - x^2/x^2} = \lim_{x \to -\infty} \frac{x + 2 + \frac{1}{x^2}}{\frac{5}{x^2} - 1}.\]

Evaluate the simplified limit

As \( x \to -\infty \), the terms \( \frac{1}{x^2} \) and \( \frac{5}{x^2} \) approach zero because they are divided by an increasingly large number. This reduces the expression to:\[\lim_{x \to -\infty} \frac{x + 2}{-1}\]Further simplifying gives \( \lim_{x \to -\infty} \left( -x - 2 \right) \). As \( x \to -\infty \), \( -x \to \infty \), so the limit approaches \( \infty \).

Key concepts

Dominant Term Identification

When evaluating limits at infinity, identifying the dominant term is crucial. This is because the dominant term is the term whose degree is the highest in the polynomial. As the variable approaches infinity or minus infinity, this term exerts the most influence on the behavior of the function. For the given expression \( \frac{x^3 + 2x^2 + 1}{5 - x^2} \), it's important to find the dominant term in both the numerator and the denominator.

• In the numerator \( x^3 + 2x^2 + 1 \), the dominant term is \( x^3 \) because it has the highest power, which is 3.


• In the denominator \( 5 - x^2 \), the dominant term is \(-x^2\) since the degree is 2 and it is negative.

By focusing on these dominant terms, we simplify the process of finding limits as \( x \) approaches infinity or minus infinity. Recognizing these key terms helps in anticipating how the expression will behave, which simplifies subsequent calculations.

Polynomial Limits

Evaluating polynomial limits involves understanding how the highest-degree terms dictate the behavior of the function as \( x \) approaches very large positive or negative values. In the context of our exercise, this means examining how the dominant terms \( x^3 \) and \(-x^2\) influence the limit.

To determine the limit, we focus primarily on the comparison between \( x^3 \) in the numerator and \(-x^2\) in the denominator. Since \( x^3 \) grows faster than \(-x^2\), it gives the direction the limit will take. Thus, we divide the entire expression by the highest power of \( x \) in the denominator. This helps us simplify the limit expression efficiently.

Ultimately, we're interested in what happens to these reduced terms as \( x \) approaches minus infinity. By simplifying further, it becomes evident what the behavior will be when the polynomial terms are dominant.

Simplification of Rational Expressions

Simplifying rational expressions is an essential step when evaluating limits, particularly as \( x \) approaches infinity or minus infinity. This simplification makes the mathematics more manageable and clarifies the evaluation of the limit. Here, we divide every term in the original rational expression by the highest power of \( x \) in the denominator, which helps in eliminating minor influences and focusing on the principal behavior.

Given the expression \( \frac{x^3 + 2x^2 + 1}{5 - x^2} \), we divide each component by \( x^2 \). This reduces the expression to \( \frac{x + 2 + \frac{1}{x^2}}{\frac{5}{x^2} - 1} \).

• The term \( \frac{1}{x^2} \) in the numerator and \( \frac{5}{x^2} \) in the denominator approach zero as \( x \to -\infty \), simplifying the overall expression.


• This leaves us with \( \frac{x + 2}{-1} \), which is easier to evaluate.

Through simplification, the problem becomes much clearer, showing how the limit behaves at infinity and leading us directly to the result.