6.2 Asymptotes and Limits
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)By the end of this section, you will be able to:
- intuitively understand the concept of the limit of a function
- understand limit notation
- describe asymptotic behavior using limits
- intuitively evaluate limits of rational functions to assess vertical, horizontal, and slant asymptotes
It's highly possible that you will spend the first chunk of your Calculus I course talking about limits, as they're a vital tool building up to the definition of a derivative, which is an elusive mythical Pokemon that perhaps you've heard of and are excited to meet. I want to give you a sneak peek at the intuitive and graphical conception of a limit, and along the way be a little more precise about vertical, horizontal, and slant asymptotes as pertains to rational functions.
When looking at a graph of \(y = f(x) \), the limit of \(f\) at \(x = a\) is the \(y\)-value that the function appears to be heading toward as you move towards a particular \(x\)-value of interest, \(a\), from both the left and the right. The existence of a limit does depend on the function actually approaching a value from both directions.
In other words, nicknaming the limit "\(L\)" if it exists, we're saying that as \(x \rightarrow a \), \(f(x) \rightarrow L \), in the arrow notation we've used before for end behavior. NOTE: the limit does not care what the actual function value \(f(a)\) is, or even whether it's defined! It's all about where the function ACTS like it's going.
We use the limit notation
\[ \lim_{x \rightarrow a} f(x) = L \notag \]
if the limit exists properly. If a function tends toward \( \pm \infty \) when approaching \( x = a \) from both directions, the limit is infinite, and although \(\infty\) is not a number, we still use the notation
\[ \lim_{x \rightarrow a} f(x) = (\pm) \infty \notag \]
Sometimes we are interested in one-sided limits, looking at the behavior of the function as we come in from only one side. In this case, we use a "\(^+\)" decoration to indicate approaching from the right, and a "\(^-\)" to indicate approaching from the left.
\[ \text{approaching \(x = a\) from the right only: } \lim_{x \rightarrow a^+} f(x) = L, \quad \text{ approaching from the left only: } \lim_{x \rightarrow a^- } f(x) = L \notag \]
The information above will make a lot more sense after looking at some pictures. Study the examples below carefully to interpret the definition.
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You can see that to find the limit as \( x \rightarrow a\), just find \(a\) on the \(x\)-axis, track it up to the function, and then track the height over to find the \(y\)-value at that point. This works great with the functions above because they're not very chaotic, there are no holes or jumps, etc. We need to make sure we grasp the fringe cases when looking for limits graphically. Study the table below.
| The limit exists and agrees with the function value at that location. | The limit exists even though the function is not defined at that location. |
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| The one-sided limits exist, but do not agree, so the limit as a whole does not exist. | The one-sided limits are infinite, but do not agree, so the limit as a whole does not exist. |
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| The function goes up forever, so the "limit at infinity" is infinite. | The function goes down forever, so the "limit at negative infinity" is negative infinity. |
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| The one-sided limits are infinite, but they do agree, so we say the limit is infinity. | The "limit at infinity" is leveling out, hugging a certain horizontal line, so it is finite. |
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You can practice reading limits off of a graph below.
Using the graph of \(y = f(x)\) below, find the limits or tell if they do not exist.
\[ \lim_{x \rightarrow -3} f(x), \quad \lim_{x \rightarrow 0^-} f(x), \quad \lim_{x \rightarrow 0^+} f(x), \quad \lim_{x \rightarrow 0} f(x), \notag \]
\[ \lim_{x \rightarrow 1} f(x), \quad \lim_{x \rightarrow 4^+} f(x), \quad \lim_{x \rightarrow 4} f(x), \quad \lim_{x \rightarrow \infty} f(x) \notag\]
- Answer
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- \( \lim_{x \rightarrow -3} f(x) = 1 \)
- \( \lim_{x \rightarrow 0^-} f(x) = 4 \)
- \( \lim_{x \rightarrow 0^+} f(x) = \infty \)
- \( \lim_{x \rightarrow 0} f(x) \) DNE.
- \( \lim_{x \rightarrow 1} f(x) = 1 \)
- \( \lim_{x \rightarrow 4^+} f(x) = -3 \)
- \( \lim_{x \rightarrow 4} f(x) \) DNE.
- \( \lim_{x \rightarrow \infty} f(x) = \infty\)
Now let's connect these ideas to the asymptotes we were learning about.
The vertical line \( x = a\) is a vertical asymptote of a function \( f(x) \) if ANY of the following are true:
| \[ \lim_{x \rightarrow a} f(x) = \infty \notag \] | \[ \lim_{x \rightarrow a^+} f(x) = \infty \notag \] | \[ \lim_{x \rightarrow a^-} f(x) = \infty \notag \] |
| \[ \lim_{x \rightarrow a} f(x) = -\infty \notag \] | \[ \lim_{x \rightarrow a^+} f(x) = -\infty \notag \] | \[ \lim_{x \rightarrow a^-} f(x) = -\infty \notag \] |
The horizontal line \(y = L \) is a horizontal asymptote of a function \(f(x)\) if
\[ \lim_{x \rightarrow \infty} f(x) = L \quad \text{ and/or } \quad \lim_{x \rightarrow -\infty} f(x) = L \notag \]
I want to do a little bit of intuitive limit-finding using algebra, which will be made more rigorous in Calc I.
Find the horizontal asymptotes of the functions by using logic and algebra to analyze \( \lim_{x \rightarrow \pm \infty} f(x) \).
1. \( f(x) = \dfrac{1}{x} \)
2. \( f(x) = \dfrac{x}{x+1} \)
3. \( f(x) = \dfrac{ x^2 + 1}{2x^2 +x + 3} \)
Solution
1. To find H.A.s, we are interested in what happens as \(x \rightarrow \infty\) and \(x \rightarrow -\infty\). Let's think about this... if \(x\) is getting bigger and bigger, what happens to \( \dfrac{1}{x}\)? Think about things like \(\frac{1}{2}, \frac{1}{4}, \frac{1}{10}, \frac{1}{10000} \)... what's happening? It's getting smaller and smaller, getting close to zero but never exactly zero. It doesn't matter that it never quite hits zero. It's always heading towards zero, so that's what we report as the limit.
\[ \lim_{x \rightarrow \infty} \frac{1}{x} = 0 \notag \]
The same thing happens as \(x \rightarrow -\infty\). Think about \( -\frac{1}{2}, -\frac{1}{10}, -\frac{1}{10000}\)... Still getting closer and closer to zero forever! We have a horizontal asymptote \( y = 0\).
2. This is a little trickier. If \(x \rightarrow \infty\), the numerator \(x\) will blow up forever, but so will the denominator \(x + 1\)! Let's imagine some possibilities as \(x\) blows up...
\[ \frac{ 2}{3}\quad ... \quad \frac{ 5}{6} \quad ... \quad \frac{20}{21} \quad ... \quad \frac{1,000}{1,001} \quad ... \quad \frac{1,000,000}{1,000,001} \notag \]
Do you agree that as \(x\) gets bigger, the \(+1\) happening in the denominator becomes basically irrelevant? These fractions are all getting closer and closer to the number 1. We have a horizontal asymptote \(y = 1\). Here's a trick to see this easily... Multiply top and bottom by \(\frac{1}{x} \):
\[ \frac{ x}{(x+1)}\frac{\cdot \frac{1}{x}}{\cdot \frac{1}{x}} = \frac{ 1}{1 + \frac{1}{x}} \notag \]
Now think about what happens when \(x \rightarrow \infty\). The \( \frac{1}{x}\) term will head towards 0, making the whole fraction head towards \( \dfrac{1}{1+0} = 1\).
3. Let's try the same trick on this one. The highest power of \(x\) in the denominator is \(x^2\), so multiply top and bottom by the reciprocal of that.
\[ \dfrac{ (x^2 + 1)}{(2x^2 +x + 3)}\frac{ \cdot \frac{1}{x^2}}{\cdot \frac{1}{x^2}} = \frac{ 1 + \frac{1}{x^2}}{2 + \frac{1}{x} + \frac{3}{x^2}} \notag \]
As \(x \rightarrow \infty\), the \(\frac{1}{x^2} \) heads to zero real quick. So does \(\frac{1}{x}\) as we saw before. And convince yourself that even though there's a 3 in \( \frac{3}{x^2} \), it will quickly be dominated by the denominator blowing up. So our whole function is heading towards
\[ \frac{ 1 + 0 }{2 + 0 + 0} = \frac{1}{2} \notag\]
We have a horizontal asymptote \(y = \frac{1}{2} \).
Find the horizontal asymptotes of the rational functions by analyzing \( \lim_{x \rightarrow \pm \infty} f(x) \).
1. \( f(x) = \dfrac{2}{x^2} \)
2. \( f(x) = \dfrac{ 4x^2 - 2}{x^2 - 9} \)
3. \( f(x) = \dfrac{1}{(x^2+1)^2} \)
- Answer
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- H.A. \( y = 0 \)
- H.A. \( y = 4 \)
- H.A. \( y = 0 \)
Tell whether \(x = 3\) is a vertical asymptote of the functions by analyzing \( \lim_{x \rightarrow 3} f(x) \).
1. \( f(x) = \dfrac{1}{x-3} \)
2. \( f(x) = \dfrac{x+2}{x+3} \)
3. \( f(x) = \dfrac{x-3}{x^2 - 9} \)
Solution
1. Let's think about what happens to \(f(x)\) if \(x\) gets close to \(3\). Consider the denominator and imagine plugging in options for \(x\) coming in from the right, numbers bigger than 3...
| \(x\)-value near \(3\) | 4 | 3.5 | 3.1 | 3.01 |
| denominator \(x - 3 \) | \(4-3 = 1\) | \( 3.5-3 = 0.5\) | \( 3.1 -3 = 0.1 \) | \( 3.01 - 3 = 0.01 \) |
Okay, we can see that the denominator is getting super tiny, heading towards 0 as \(x\) heads towards 3. If the denominator of a fraction gets smaller, the fraction as a whole blows up! So we have \(x = 3\) as a V.A. because
\[ \lim_{x \rightarrow 3^+} f(x) = \infty \notag \]
2. Thinking about \( x\) becoming 3, we see the numerator will head towards \( 3+2 = 5\) and the denominator will head towards \( 3 + 3 = 6 \). Nothing special happening here, the fraction as a whole is heading towards \( \frac{5}{6} \). That's a finite number, not \( \pm \infty\), so we do not have a vertical asymptote here.
3. If \(x \rightarrow 3 \), we see that both our numerator and denominator are heading towards 0, so we don't know what the fraction is doing as a whole... Let's try simplifying this expression to see if anything becomes clearer.
\[ \frac{x-3}{x^2 - 9} = \frac{ x-3}{(x+3)(x-3)} = \frac{1}{x+3} \notag \]
Aha! This function \(f\) isn't defined at \(x = 3 \), but it behaves exactly like \( \dfrac{1}{x+3} \). When \(x \rightarrow 3\), that expression heads toward \( \frac{1}{9} \). In this case, \(x = 3\) is a hole in the graph, but not a vertical asymptote!
Tell whether \(x = 0\) is a vertical asymptote of the functions by analyzing \( \lim_{x \rightarrow 0} f(x) \).
- \( f(x) = \dfrac{ 1}{x^2} \)
- \( f(x) = \dfrac{ x}{x+1} \)
- \( f(x) = \dfrac{ x^2 + x}{x} \)
- \( f(x) = \dfrac{ x+1}{x^2} \)
- Answer
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- Yes, because \( \lim_{x \rightarrow 0} f(x) = \infty \).
- No, \( \lim_{x \rightarrow 0} \dfrac{x}{x+1} = \dfrac{ 0}{1} = 0 \).
- Simplifying the expression, \(f\) behaves like \( x+1 \), and \( \lim_{x \rightarrow 0}(x+1) = 1 \), so no.
- Yes, as \(x \rightarrow 0\), the numerator heads to \(1\), but the denominator gets very tiny very quickly, making the whole fraction blow up to \(\infty\).
Slant Asymptotes
I want to use limits to talk about one final concept relating to rational functions. Sometimes a rational function appears to do some crazy stuff, but then hug a straight line going off forever to infinity, like the pictures below:
What's going on here is that, as \(x \rightarrow \pm \infty\), the function ends up behaving like a linear function. Look at the first example. Notice what happens if I start trying to analyze the limit as \(x \rightarrow \infty\).
\[ \lim_{x \rightarrow \infty} \left( \frac{ x^2 + 2x +1}{x} \right) = \lim_{x \rightarrow \infty} \left( \frac{ x^2}{x} + \frac{2x}{x} +\frac{1}{x} \right) = \lim_{x \rightarrow \infty} \left( x + 2 + \frac{1}{x} \right) \notag \]
Hmm... As \(x\) blows up, that \( \frac{1}{x} \) term is going to contribute less and less to the behavior of the function. My graph will end up acting just like the first piece, a straight line, \(y = x+2\).
A rational function \(f(x) = \dfrac{p(x)}{d(x)} \) has a slant asymptote when the degree of the numerator is exactly one more than the degree of the denominator. To find the equation of the slant asymptote line, perform polynomial long division on \( \dfrac{p(x)}{d(x)} \), writing \( \dfrac{p(x)}{d(x)} = q(x) + \dfrac{r(x)}{d(x)} \). The quotient \(q(x)\) will be a linear function, and the \( \dfrac{r(x)}{d(x)} \) will become irrelevant as \(x \rightarrow \infty\). Then the slant asymptote is \( y = q(x) \).
Find the slant asymptote of the second example above, \(f(x) = \dfrac{ 2x^3 + 2x + 1}{x^2 - 1} \).
Solution
Polynomial long division yields \( f(x) = (2x) + \dfrac{ 4x-1}{x^2 -1} \). That second term will eventually become tiny as \(x \rightarrow \infty\), which we can see by doing that trick for analyzing infinite limits:
\[ \dfrac{( 4x-1)}{(x^2 -1)} \dfrac{ \cdot \frac{1}{x^2}}{\cdot \frac{1}{x^2}} = \dfrac{ \frac{4}{x} - \frac{1}{x^2}}{1 - \frac{1}{x^2}} \notag \]
The numerator heads to 0, while the denominator heads to 1. Well, \( \frac{0}{1} = 0\), so we see that term will become so tiny it's irrelevant! Thus \(f\) will act like \(q(x) = 2x \), a line.
Find the slant asymptotes of the rational functions below. Confirm afterwards by looking at the graph.
| \( f(x) = \dfrac{ x^2}{x+1} \) | \( f(x) = - \dfrac{x^3+x^2+x+1}{x^2} \) |
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- Answer
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- Long division yields \( (x-1) + \frac{1}{x+1} \). The slant asymptote is \(y = x-1\).
- Long division (or just splitting the fraction and cancelling) yields \( - \left(x+1 + \frac{1}{x} + \frac{1}{x^2} \right) \). The last two terms approach zero as \(x \rightarrow \infty\), so the function hugs the slant asymptote \( y = -x -1 \).