Question

Find the horizontal asymptotes of each function using limits at infinity. $$f(x)=\frac{3 e^{5 x}+7 e^{6 x}}{9 e^{5 x}+14 e^{6 x}}$$

Short answer

Answer: The horizontal asymptote is \(y=\frac{1}{2}\) as x approaches positive infinity.

Step-by-step solution

Identify the highest power of x

In the given function, \(f(x)=\frac{3 e^{5 x}+7 e^{6 x}}{9 e^{5 x}+14 e^{6 x}}\), the highest power of x is \(e^{6x}\), which appears in both the numerator and denominator.

Divide by the highest power of x

Divide both the numerator and the denominator by \(e^{6x}\), which will simplify the function to find the limit: $$g(x)=\frac{\frac{3 e^{5 x}}{e^{6x}}+\frac{7 e^{6 x}}{e^{6x}}}{\frac{9 e^{5 x}}{e^{6x}}+\frac{14 e^{6 x}}{e^{6x}}} = \frac{3e^{-x} +7}{9e^{-x} +14} $$

Find the limits as x approaches infinity

Now, we will analyze the limits of \(g(x)\) as x approaches positive and negative infinity: $$\lim_{x\to +\infty} \frac{3e^{-x} +7}{9e^{-x} +14}$$ $$\lim_{x\to -\infty} \frac{3e^{-x} +7}{9e^{-x} +14}$$ As x approaches positive infinity, \(e^{-x}\) approaches 0, and as x approaches negative infinity, \(e^{-x}\) approaches infinity.

Calculate the limits

Calculate the limits as x approaches positive and negative infinity: $$\lim_{x\to +\infty} \frac{3e^{-x} +7}{9e^{-x} +14} = \frac{3(0) +7}{9(0) +14} = \frac{7}{14} = \frac{1}{2}$$ $$\lim_{x\to -\infty} \frac{3e^{-x} +7}{9e^{-x} +14} = \text{indeterminate form}\ (0/0)$$ Since the limit at negative infinity is in indeterminate form, we cannot conclude that there is a horizontal asymptote at negative infinity.

Identify the horizontal asymptotes

Thus, the function \(f(x)=\frac{3 e^{5 x}+7 e^{6 x}}{9 e^{5 x}+14 e^{6 x}}\) has a horizontal asymptote at \(y=\frac{1}{2}\) as \(x\) approaches positive infinity.

Key concepts

Limits at Infinity

Understanding the behavior of functions as the input values become very large is essential in mathematics. Limits at infinity refer to what value a function approaches as the variable approaches positive or negative infinity.

In the given exercise, we look at the limit of the function \(f(x)=\frac{3 e^{5 x}+7 e^{6 x}}{9 e^{5 x}+14 e^{6 x}}\) as \(x\) approaches infinity. Since the highest powers of \(e^x\) are the dominant terms when \(x\) is very large, they determine the function's end-behavior.

To simplify the process, division by the highest power of \(e^x\) in both the numerator and the denominator is done, leading to a new function which is easier to analyze. Finally, evaluating the limit as \(x\) approaches infinity helps us find the horizontal asymptote, which indicates the value the function is getting closer to, without actually reaching it.

In practice, if after simplification the highest powers of \(e^x\) cancel out, and if the remaining highest-degree terms in the numerator and the denominator are constants, their ratio will determine the horizontal asymptote, as seen with the resultant \(\frac{1}{2}\) in this exercise.

Exponential Functions

Exponential functions are powerful tools in mathematics, characterized by their variable exponents. The general form is \(f(x) = a^{x}\), where \(a\) is a constant and \(x\) is an exponent. In our context, the exponential functions are expressed in terms of Euler's number, \(e\), which is approximately equal to 2.71828.

An important property of exponential functions like \(e^x\) is that they grow or decay very rapidly. In the function from our exercise, we have terms like \(e^{5x}\) and \(e^{6x}\), which indicate exponential growth and decay, respectively, when \(x\) is positive or negative.

As \(x\) becomes very large, \(e^{-x}\) approaches zero because the function represents rapid decay. This property is crucial in determining the horizontal asymptote as it essentially causes the terms containing \(e^{-x}\) to vanish when considering the limit at positive infinity. This helps simplify many complex-looking functions into more manageable forms for analysis.

Indeterminate Forms

In calculus, indeterminate forms occur when evaluating limits and the expression does not clearly indicate a specific value. Common examples include \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\), which came up in this problem when attempting to evaluate the limit as \(x\) approaches negative infinity.

An indeterminate form is a sign that further analysis is required; it does not mean the limit does not exist. To deal with this, mathematicians have developed tools such as L'Hôpital's rule, algebraic simplification, or factorization to resolve these forms and find a precise limit.

In the exercise, the calculated limit as \(x\) approaches negative infinity resulted in an indeterminate form, indicating that the horizontal asymptote cannot be concluded from this directly. However, since exponential terms grow without bound, the coefficients at powers of \(e\) become irrelevant, simplifying the limit evaluation process.