Problem 13 - Precalculus: Mathematical for Calculus

Question

Find the limit. $$\lim _{x \rightarrow-\infty}\left(\frac{x-1}{x+1}+6\right)$$

Step-by-Step Solution

Answer

The limit is 7.

Step 1: Rewrite the Original Limit Problem

We need to find \( \lim _{x \rightarrow -\infty} \left( \frac{x-1}{x+1} + 6 \right) \). Our goal is to simplify the expression inside the limit so we can evaluate it as \( x \) approaches \( -\infty \).

Step 2: Simplify the Fraction

Simplify \( \frac{x-1}{x+1} \) by dividing both the numerator and the denominator by \( x \). This gives us: \[ \frac{x/x - 1/x}{x/x + 1/x} = \frac{1 - \frac{1}{x}}{1 + \frac{1}{x}}. \] As \( x \rightarrow -\infty \), \( \frac{1}{x} \rightarrow 0 \).

Step 3: Evaluate the Limit of the Simplified Fraction

Given \[ \lim_{x \rightarrow -\infty} \frac{1 - \frac{1}{x}}{1 + \frac{1}{x}} = \frac{1 - 0}{1 + 0} = 1, \] we can substitute this result back into the expression for the original limit.

Step 4: Calculate the Final Limit

Now substitute back into the limit expression: \( \lim_{x \rightarrow -\infty} \left( \frac{x-1}{x+1} + 6 \right) = 1 + 6 = 7.\)

Step 5: Conclusion

The limit \( \lim_{x \rightarrow -\infty} \left( \frac{x-1}{x+1} + 6 \right) = 7.\)

Key Concepts

Infinite Limits

In calculus, infinite limits involve functions where the value being approached is infinity. This concept comes into play when we evaluate how a function behaves as the input value, denoted by \( x \), heads towards positive or negative infinity. For example, when we see notation like \( \lim_{x \rightarrow -\infty} \), it asks what happens to the function's value as \( x \) approaches incredibly large negative numbers. Infinite limits often arise in rational functions or exponential expressions. Understanding them helps us predict behavior without graphing the function every time.A key part of solving infinite limits is recognizing the behavior of different function components as \( x \) grows or shrinks infinitely. Some parts may diminish to zero, while others stabilize to a value or grow infinitely. Simplifying the expression wisely will illuminate these behaviors.

Rational Functions

Rational functions are fractions where both the numerator and the denominator are polynomials. An example looks like \( \frac{x-1}{x+1} \). These types of functions frequently appear in calculus when dealing with limits and asymptotic behavior.Understanding the structure of rational functions is crucial, especially when simplifying expressions. Common techniques involve factoring the polynomials or dividing through by the highest power of \( x \) present in the denominator. This helps reveal how different terms behave as \( x \) approaches infinity. For instance, if you divide each term in \( \frac{x-1}{x+1} \) by \( x \), the expression simplifies to \( \frac{1-\frac{1}{x}}{1+\frac{1}{x}} \). This form makes it easier to see that as \( x \rightarrow -\infty \), \( \frac{1}{x} \rightarrow 0 \), which makes the fraction approach a specific constant, rather than tending towards another polynomial or infinity.

Limit Evaluation

Limit evaluation is about understanding what value a function approaches as the input approaches a certain number or infinity. Often, this involves algebraic manipulation to simplify the function to a point where its behavior becomes clear.To start evaluating a limit, identify any indeterminate forms or terms that become negligible as \( x \) grows infinitely large or small. Factoring, multiplying by conjugates, or using the sandwich theorem are common techniques.In our example, \( \lim_{x \rightarrow -\infty} \left( \frac{x-1}{x+1} + 6 \right) \), we simplify the fraction \( \frac{x-1}{x+1} \). This fraction transformed to \( \frac{1-\frac{1}{x}}{1+\frac{1}{x}} \) allows us to see the impact of large values of \( x \). At this stage, observing \( \frac{1}{x} \rightarrow 0 \) simplifies the expression further to \( 1 \). Finally, adding the constant (in this case, 6), completes our evaluation to get \( 7 \), making this a clear process of assessing how values tend to behave.