Question

In Exercises \(21-26,\) find \(\lim _{x \rightarrow \infty} y\) and \(\lim _{x \rightarrow-\infty} y\). $$y=\left(2-\frac{x}{x+1}\right)\left(\frac{x^{2}}{5+x^{2}}\right)$$

Short answer

\(\lim_{x\rightarrow\infty} y = 1\) and \(\lim_{x\rightarrow-\infty} y = 1\)

Step-by-step solution

Simplifying the function

Simplify each fraction in the function separately. First, simplify \(\frac{x}{x+1}\). As \(x\) approaches ±∞, the +1 becomes insignificant, thus the term simplifies to \(\frac{x}{x}\) which equals 1. Simplify \(\frac{x^{2}}{5+x^{2}}\). Again, as \(x\) approaches ±∞, the +5 becomes insignificant, thus that term simplifies to \(\frac{x^{2}}{x^{2}}\) which equals 1. Substitute back into the function to get a simplified function: \(y=(2-1)(1)=1\)

Calculate the limit of the function

Substitute the value of the function into the limit formula: \(\lim_{x\to\infty} y = 1\) and \(\lim_{x\to-\infty} y = 1\). Because no matter the value of \(x\), the function \(y\) maintains a constant value of 1.

Key concepts

Limit of a Function

Understanding the limit of a function is foundational in Calculus. It deals with determining the behavior of a function as its input approaches a particular value. In the context of infinity, we are interested in what happens to the function's output as the input grows without bound, either positively (toward +∞) or negatively (toward -∞).

For example, when we analyze the limit of the function \( y = (2 - \frac{x}{x+1})\left(\frac{x^2}{5+x^2}\right) \) as \( x \) approaches infinity, we're seeking to know what value \( y \) gets closer to. In this case, both \( \frac{x}{x+1} \) and \( \frac{x^2}{5+x^2} \) approach 1, therefore simplifying the expression to \( y = (2-1)(1) = 1 \). The limit as \( x \) tends to both +∞ and -∞ is 1, implying that no matter how large or small \( x \) gets, the value of \( y \) will approach 1.

Asymptotic Behavior

The concept of asymptotic behavior is closely related to limits at infinity. It describes how a function behaves as it moves closer and closer to a line called an asymptote. Asymptotes can be vertical, horizontal, or even oblique (diagonal), but in the context of limits at infinity, we're typically dealing with horizontal asymptotes.

When we observe that the limit of \( y = (2 - \frac{x}{x+1})\left(\frac{x^2}{5+x^2}\right) \) is 1 for both \( x \) tending to +∞ and -∞, we understand that the horizontal line \( y = 1 \) is a horizontal asymptote of the function. That's to say, as \( x \) grows indefinitely in either the positive or negative direction, the function will increasingly resemble the line \( y = 1 \) but never actually reach it.

Simplifying Complex Fractions

Dealing with complex fractions, especially those involving variables, can be daunting. But simplifying complex fractions is a critical skill, especially when analyzing limits at infinity. When tackling such fractions, look for dominant terms—these are the highest-power terms in the numerator and the denominator, which have the most significant influence on the value of the fraction as the variables grow large.

For instance, in the fraction \( \frac{x}{x+1} \), the dominant term is \( x \), because as \( x \) becomes very large, the +1 has less and less impact. Similarly, in \( \frac{x^2}{5+x^2} \), \( x^2 \) is the dominant term in both the numerator and the denominator. This simplification principle lets us reduce these fractions to 1 when \( x \) approaches infinity, leading to the function simplifying to \( y = 1 \), a constant value.

Infinite Limits

While some limits involve approaching a specific finite number, infinite limits concern a function's behavior as it approaches infinite values of the input. Not to be confused with limits that result in infinity, infinite limits question what finite value a function is heading towards when the input endlessly increases or decreases in magnitude.

In our example with \( y = (2 - \frac{x}{x+1})\left(\frac{x^2}{5+x^2}\right) \), we calculated the infinite limits by determining the value that \( y \) approaches as \( x \) approaches positive and negative infinity. Despite the input growing without bounds, the output converges to a fixed number, in this case, the limit \( y = 1 \). This demonstrates that even as \( x \) increases or decreases without limit, the function can still tend toward a constant value, characterized by its infinite limits.