Problem 63 - Precalculus

Question

For Exercises 61 through \(64,\) evaluate the limits by dividing the numerator and denominator by the highest power of \(x\) occurring in the denominator. $$\lim _{x \rightarrow-\infty} \frac{2 x^{2}-1}{x^{3}+2 x+12}$$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Identify Highest Power

In the expression \( \frac{2x^2 - 1}{x^3 + 2x + 12} \), the highest power of \( x \) in the denominator is \( x^3 \). We'll use this highest power \( x^3 \) to simplify the expression.

Step 2: Divide Each Term

Divide every term in both the numerator and the denominator by \( x^3 \). This will help simplify the limit expression:\[\lim_{x \to -\infty} \frac{\frac{2x^2}{x^3} - \frac{1}{x^3}}{\frac{x^3}{x^3} + \frac{2x}{x^3} + \frac{12}{x^3}}\] which simplifies to \[\lim_{x \to -\infty} \frac{\frac{2}{x} - \frac{1}{x^3}}{1 + \frac{2}{x^2} + \frac{12}{x^3}}\].

Step 3: Evaluate the Limit

As \( x \to -\infty \), each term with \( x \) in the denominator approaches zero:\[\lim_{x \to -\infty} \frac{\frac{2}{x} - \frac{1}{x^3}}{1 + \frac{2}{x^2} + \frac{12}{x^3}} = \lim_{x \to -\infty} \frac{0 - 0}{1 + 0 + 0} = 0.\]

Key Concepts

Dividing by Highest Power of x

One of the essential techniques in evaluating limits at infinity, particularly for rational functions, is dividing the numerator and the denominator by the highest power of \( x \) found in the denominator. This method simplifies the expression by minimizing the influence of lower power terms as \( x \) approaches infinity or negative infinity.

Here's how it works:

• Identify the highest power of \( x \) in the denominator. This is crucial because rational functions often become undefined at infinity unless simplified.
• Divide every term in the numerator and the denominator by this highest power of \( x \). This operation transforms terms with smaller exponentials of \( x \) towards zero.

By reducing the expression to terms that either vanish or simplify significantly, you make it easier to assess the behavior of the function as \( x \) approaches infinity or negative infinity. This process is important because it provides clarity where simplification by conventional methods isn't straightforward.

Evaluating Limits at Infinity

Evaluating limits as \( x \) approaches infinity or negative infinity is a fundamental skill in calculus. It helps us understand the asymptotic behavior of functions. Here's a straightforward approach:

• After dividing by the highest power of \( x \), most terms will reduce to fractions with increasing powers of \( x \) in the denominator, which approach zero as \( x \) becomes very large or very small.
• Focus on the terms that do not reduce to zero, determine their final contribution to the limit.

For example, if you have simplified a rational function to an expression like \( \frac{0 + 0}{1 + 0 + 0} \), the limit evaluates to simply the constant remaining in the denominator, which is 1 in this case. Thus, the computed limit is zero if the numerator simplifies entirely to zero. This method systematically assesses each component to ascertain the final outcome without ambiguity.

Rational Functions Limits

Rational functions, which are ratios of polynomial functions, often pose challenges in limit evaluation particularly at infinity. However, employing strategic simplification can turn complex expressions into straightforward problems.

• Start by identifying the variable terms that dominate the behavior of the function as \( x \) approaches infinity or negative infinity.
• Applying the technique of dividing by the highest power of \( x \) helps isolate these dominating terms.

Once dominant terms are identified, the reduced function should display a clear pathway to understanding its limit based on the expression left behind after division. Commonly, this leads to expressions where most terms vanish, leaving behind a simple result, such as zero or another constant, which conclusively characterizes the function's behavior at infinity. This systematic approach demystifies the evaluation process, making rational functions much easier to handle for students.