Question

Use the following instructions to determine the end behavior of \(f(x)=\frac{e^{x}+e^{2 x}}{e^{2 x}+e^{3 x}}\) a. Evaluate \(\lim _{x \rightarrow \infty} f(x)\) by first dividing the numerator and denominator by \(e^{3 x}\). b. Evaluate \(\lim _{x \rightarrow-\infty} f(x)\) by first dividing the numerator and denominator by \(e^{2 x}\). c. Give the horizontal asymptote(s). d. Graph \(f\) to confirm your work in parts (a)-(c).

Short answer

Answer: The end behavior of the function is: 1. As \(x\) approaches infinity, \(f(x)\) approaches 0 and has a horizontal asymptote at \(y=0\). 2. As \(x\) approaches negative infinity, \(f(x)\) approaches infinity, and there is no finite horizontal asymptote.

Step-by-step solution

Divide the numerator and denominator by \(e^{3x}\)

We divide both the numerator and denominator of the function by \(e^{3x}\): $$ \frac{e^{x}+e^{2 x}}{e^{2 x}+e^{3 x}}\cdot\frac{\frac{1}{e^{3x}}}{\frac{1}{e^{3x}}}=\frac{e^{x}e^{-3x}+e^{2x}e^{-3x}}{e^{2x}e^{-3x}+e^{3x}e^{-3x}}=\frac{e^{-2x}+e^{-x}}{e^{-x}+1} $$

Calculate the limit as x approaches infinity

Now we take the limit as x approaches infinity: $$ \lim_{x\rightarrow\infty}\frac{e^{-2x}+e^{-x}}{e^{-x}+1} $$ As \(x\) approaches infinity, \(e^{-x}\) and \(e^{-2x}\) both approach 0. Therefore, the limit is: $$ \lim_{x\rightarrow\infty}\frac{e^{-2x}+e^{-x}}{e^{-x}+1}=\frac{0+0}{0+1} = 0 $$ #b. Evaluate \(\lim _{x \rightarrow-\infty} f(x)\) by first dividing the numerator and denominator by \(e^{2 x}\).#

Divide the numerator and denominator by \(e^{2x}\)

We divide both the numerator and denominator of the function by \(e^{2x}\): $$ \frac{e^{x}+e^{2 x}}{e^{2 x}+e^{3 x}}\cdot\frac{\frac{1}{e^{2x}}}{\frac{1}{e^{2x}}}=\frac{e^{x}e^{-2x}+e^{2x}e^{-2x}}{e^{2x}e^{-2x}+e^{3x}e^{-2x}}=\frac{e^{-x}+1}{1+e^{x}} $$

Calculate the limit as x approaches negative infinity

Now we take the limit as x approaches negative infinity: $$ \lim_{x\rightarrow-\infty}\frac{e^{-x}+1}{1+e^{x}} $$ As \(x\) approaches negative infinity, \(e^{-x}\) approaches infinity, and \(e^x\) approaches 0. Therefore, the limit is: $$ \lim_{x\rightarrow-\infty}\frac{e^{-x}+1}{1+e^{x}}=\frac{\infty+1}{1+0} = \infty $$ #c. Give the horizontal asymptote(s).#

Horizontal Asymptote(s)

Using the results from our limits, we have two horizontal asymptotes: 1. As \(x\) approaches infinity, \(f(x)\) approaches 0, so \(y=0\) is a horizontal asymptote. 2. As \(x\) approaches negative infinity, \(f(x)\) approaches infinity; there is no finite horizontal asymptote for this case. #d. Graph \(f\) to confirm your work in parts (a)-(c).# For this step, please use a graphing tool (such as Desmos, GeoGebra, or a graphing calculator) to plot the function: $$ f(x)=\frac{e^{x}+e^{2 x}}{e^{2 x}+e^{3 x}} $$ Observe the graph and confirm that it has the horizontal asymptotes \(y=0\) (as \(x\) approaches infinity) and no finite horizontal asymptote as \(x\) approaches negative infinity.

Key concepts

Limits at Infinity

Understanding the concept of limits at infinity is crucial for analyzing the end behavior of functions. A limit at infinity seeks the value that a function approaches as the variable moves towards positive or negative infinity. In the example of the function\( f(x)=\frac{e^{x}+e^{2 x}}{e^{2 x}+e^{3 x}} \), we are interested in two key limits: \( \lim_{x\rightarrow\infty} f(x) \) and \( \lim_{x\rightarrow-\infty} f(x) \).

To ascertain the limit at positive infinity, we observe that the exponential components \( e^{-x} \) and \( e^{-2x} \) dissipate to zero, hence the function narrows down to a limit of zero. In contrast, as x wends its way to negative infinity, the terms invert, with the effect of enlarging the function towards an infinite limit. Thus, the end behavior is determined by seeing how these expressions evolve as x becomes very large or very small, allowing us to identify the values that the function gravitates towards, or its limits at infinity.

Horizontal Asymptotes

A horizontal asymptote is a horizontal line that the graph of a function approaches but never quite reaches as the independent variable heads towards infinity or negative infinity. This concept plays a pivotal role in predicting the long-term behavior of functions. Referring to our function \( f(x) \), we observed two scenarios. First, as \( x \) tends toward positive infinity, \( f(x) \) approximates to zero, manifesting a horizontal asymptote at \( y = 0 \). This asymptote acts like a boundary that the graph strives to reach horizontally.

On the flip side, as \( x \) barrels towards negative infinity, the function rockets without bound and hence doesn't level off to reveal a horizontal asymptote. Due to this, only the horizontal asymptote at \( y = 0 \) is acknowledged. Recognizing horizontal asymptotes equips us with valuable insight into the function's end behavior over the vast expanses of the x-axis.

Exponential Functions

Exponential functions yield some of the most interesting aspects when golfing for limit behaviors. These functions, typically taking the form \( e^{kx} \) where \( k \) is a constant, morph rapidly and are exceptionally responsive to changes in the power of \( e \). Within the framework of our function \( f(x) \), exponential terms dictate the output as we surge towards infinity or plummet into negative infinity.

As we propel \( x \) towards positive infinity in exponential functions, the outputs shrink steadfastly to zero if our exponent bears a negative sign. This property cemented the horizontal asymptote at \( y = 0 \) for \( f(x) \) as \( x \) approached positive infinity. Conversely, if the exponent is positive, as \( x \) retreats into negative infinity, our function soars to larger and larger values. It's the fundamental nature of these exponential expressions that shapes the ubiquitous growth as we scrutinize their trail into the extremes of the x-domain.