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Evaluate \(\lim _{x \rightarrow \infty} f(x)\) and \(\lim _{x \rightarrow-\infty} f(x)\) for the following functions. Then give the horizontal asymptote(s) of \(f\) (if any). $$f(x)=\frac{4 x^{3}}{2 x^{3}+\sqrt{9 x^{6}+15 x^{4}}}$$

Short Answer

Expert verified
Answer: The horizontal asymptote of the function is \(y = 2\).

Step by step solution

01

Simplify the function

First, let's simplify the function: $$f(x)=\frac{4 x^{3}}{2 x^{3}+\sqrt{9 x^{6}+15 x^{4}}}$$ We can factor out \(x^3\) from the numerator and some terms of the denominator: $$f(x)=\frac{x^3(4)}{x^3(2+\frac{\sqrt{9 x^{6}+15 x^{4}}}{x^3})}$$ Now, we can cancel out the \(x^3\) terms: $$f(x)=\frac{4}{2+\frac{\sqrt{9 x^{6}+15 x^{4}}}{x^3}}$$
02

Find \(\lim _{x \rightarrow \infty} f(x)\)

As x approaches infinity, the term \(\frac{15 x^{4}}{x^3}\) becomes insignificant compared to \(\frac{9 x^{6}}{x^3}\), which simplifies to \(9x^3\). Therefore, as \(x \rightarrow \infty\), the function becomes: $$f(x) \approx \frac{4}{2+\frac{\sqrt{9 x^{6}}}{x^3}} = \frac{4}{2+\sqrt{9 x^{3}}} \rightarrow \frac{4}{2+0} = 2$$ So, \(\lim _{x \rightarrow \infty} f(x) = 2\)
03

Find \(\lim _{x \rightarrow-\infty} f(x)\)

As x approaches negative infinity, the term \(\frac{15 x^{4}}{x^3}\) also becomes insignificant compared to \(\frac{9 x^{6}}{x^3}\), which simplifies to \(9x^3\). Since \(x^3\) becomes negative as x approaches negative infinity, we will have: $$f(x) \approx \frac{4}{2+\frac{\sqrt{9 (-x^{3})}}{(-x^3)}} = \frac{4}{2-\sqrt{9 x^{3}}} \rightarrow \frac{4}{2-0} = 2$$ So, \(\lim _{x \rightarrow-\infty} f(x) = 2\)
04

Determine the horizontal asymptote(s)

Now that we have the limits, we can determine the horizontal asymptote(s) of the function: Since, both the limits at infinity and negative infinity are equal to 2, there is a horizontal asymptote at \(y=2\). In conclusion, the horizontal asymptote of the function \(f(x)\) is \(y = 2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Limits at Infinity
In calculus, when we talk about limits at infinity, we are examining how a function behaves as the input variable, usually denoted as \(x\), approaches infinity or negative infinity. These types of limits help us understand the long-term behavior of functions, especially polynomials and rational functions.
For example, consider the function \(f(x)=\frac{4 x^{3}}{2 x^{3}+\sqrt{9 x^{6}+15 x^{4}}}\). As \(x\) increases to very large positive or negative values, the limit helps us determine the end behavior of the function.
In this case, simplifying the function to consider only the dominant terms, we can find that:
  • \(\lim_{x \rightarrow \infty} f(x) = 2\)
  • \(\lim_{x \rightarrow -\infty} f(x) = 2\)
"These limits tell us that the function approaches a horizontal line at \(y=2\), no matter which direction \(x\) heads."
Understanding these limits, especially at infinity, provides crucial insight into how a function behaves as \(x\) moves towards exceptionally large or small values. This is fundamental in figuring out whether the function "levels off" or keeps growing/shrinking indefinitely.
Asymptotic Behavior
The concept of asymptotic behavior refers to how functions behave as they approach certain boundaries or infinitely large values of \(x\). An asymptote is a line that a graph of a function approaches but never touches or intersects. In this context, horizontal asymptotes are especially important for functions that "stabilize" at a certain \(y\)-value as \(x\) reaches extremes.
For the function \(f(x)=\frac{4 x^{3}}{2 x^{3}+\sqrt{9 x^{6}+15 x^{4}}}\), we have determined that:
  • As \(x\) approaches \(\infty\) or \(-\infty\), the function approaches \(y=2\).
  • Therefore, the horizontal asymptote is \(y = 2\).
Due to the dominance of certain terms as \(x\) increases or decreases without bound, the function stabilizes around this horizontal line.
Recognizing these behaviors allows mathematicians and students to predict the general shape and trend of complex functions without needing to plot numerous points.
Rational Functions
Rational functions are fractions in which both the numerator and the denominator are polynomials. The general form can be expressed as \(f(x) = \frac{P(x)}{Q(x)}\), where \(P(x)\) and \(Q(x)\) are both polynomial expressions.
These functions often have interesting traits such as vertical and horizontal asymptotes, which are influenced by the degrees of the polynomials involved. In the function \(f(x)=\frac{4 x^{3}}{2 x^{3}+\sqrt{9 x^{6}+15 x^{4}}}\), we can observe:
  • Both the numerator and denominator are dominated by \(x^3\) terms.
  • This balance leads to a well-defined horizontal asymptote at \(y = 2\).
By canceling similar terms, we simplify the function's behavior at extreme \(x\) values, showing why rational functions are significant in studying asymptotic behavior.
Rational functions are essential in modeling real-world phenomena, such as calculating growth or decay, due to their predictable asymptotic behaviors.

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Most popular questions from this chapter

Determine whether the following statements are true and give an explanation or counterexample. a. The graph of a function can never cross one of its horizontal asymptotes. b. A rational function \(f\) can have both \(\lim _{x \rightarrow \infty} f(x)=L\) and \(\lim _{x \rightarrow-\infty} f(x)=\infty\). c. The graph of any function can have at most two horizontal asymptotes.

If a function \(f\) represents a system that varies in time, the existence of \(\lim _{t \rightarrow \infty} f(t)\) means that the system reaches a steady state (or equilibrium). For the following systems, determine if a steady state exists and give the steady-state value. The amplitude of an oscillator is given by \(a(t)=2\left(\frac{t+\sin t}{t}\right)\).

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Use the following definitions. Assume fexists for all \(x\) near a with \(x>\) a. We say that the limit of \(f(x)\) as \(x\) approaches a from the right of a is \(L\) and write \(\lim _{x \rightarrow a^{+}} f(x)=L,\) if for any \(\varepsilon>0\) there exists \(\delta>0\) such that $$ |f(x)-L|<\varepsilon \quad \text { whenever } \quad 00\) there exists \(\delta>0\) such that $$ |f(x)-L|<\varepsilon \quad \text { whenever } \quad 0

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