1.4 Limits Involving Infinity
What happens far out on the $x$-axis — horizontal asymptotes, vertical asymptotes, and functions that grow without bound.
We begin by examining what it means for a function to have a finite limit at infinity. Then we study the idea of a function with an infinite limit at infinity. Earlier we met vertical asymptotes; in this section we also deal with horizontal asymptotes.
Limits at Infinity and Horizontal Asymptotes
Recall that $\displaystyle\lim_{x\to a}f(x)=L$ means $f(x)$ becomes arbitrarily close to $L$ as long as $x$ is sufficiently close to $a$. We can extend this idea to limits at infinity. Consider $f(x)=2+\dfrac1x$. As the values of $x$ get larger, the values of $f(x)$ approach $2$ (Figure 1.7, Table 1.4). We write $\displaystyle\lim_{x\to\infty}f(x)=2$. Similarly, for $x<0$, as $|x|$ gets larger, $f(x)$ approaches $2$, and we write $\displaystyle\lim_{x\to-\infty}f(x)=2$.
| $x$ | $10$ | $100$ | $1{,}000$ | $10{,}000$ |
|---|---|---|---|---|
| $2+\frac1x$ | $2.1$ | $2.01$ | $2.001$ | $2.0001$ |
| $x$ | $-10$ | $-100$ | $-1000$ | $-10{,}000$ |
| $2+\frac1x$ | $1.9$ | $1.99$ | $1.999$ | $1.9999$ |
Table 1.4 — values of $f$ as $x\to\pm\infty$. Pick “$2+1/x$” in Interactive 1 to see Figure 1.7 come alive.
More generally, $\displaystyle\lim_{x\to\infty}f(x)=L$ if $f(x)$ becomes arbitrarily close to $L$ as long as $x$ is sufficiently large; and $\displaystyle\lim_{x\to-\infty}f(x)=L$ if $f(x)$ becomes arbitrarily close to $L$ as long as $x<0$ and $|x|$ is sufficiently large.
📘 Definition — Limit at infinity
If the values of $f(x)$ become arbitrarily close to $L$ as $x$ becomes sufficiently large, we say $f$ has a limit at infinity and write $\displaystyle\lim_{x\to\infty}f(x)=L$. If the values of $f(x)$ become arbitrarily close to $L$ for $x<0$ as $|x|$ becomes sufficiently large, we say $f$ has a limit at negative infinity and write $\displaystyle\lim_{x\to-\infty}f(x)=L$. In either case the graph of $f$ approaches the line $y=L$, a horizontal asymptote of $f$.
📘 Definition — Horizontal asymptote
If $\displaystyle\lim_{x\to\infty}f(x)=L$ or $\displaystyle\lim_{x\to-\infty}f(x)=L$, the line $y=L$ is a horizontal asymptote of $f$. A function can have different horizontal asymptotes at the two ends — $y=L$ as $x\to\infty$ and $y=M$ as $x\to-\infty$. For example, $\displaystyle\lim_{x\to\infty}\frac1x=0$, so $y=0$ is a horizontal asymptote of $f(x)=\dfrac1x$.
📘 Definition — Vertical asymptote
The line $x=a$ is a vertical asymptote of the graph of $y=f(x)$ if at least one of the following is true:
$$\lim_{x\to a^-}f(x)=\pm\infty,\qquad \lim_{x\to a^+}f(x)=\pm\infty.$$
For example, for $f(x)=\dfrac{x}{x-1}$ the numerator approaches $1$ and the denominator approaches $0$ as $x\to1$, so
$$\lim_{x\to1^+}\frac{x}{x-1}=+\infty,\qquad\lim_{x\to1^-}\frac{x}{x-1}=-\infty,$$
and the curve has the vertical asymptote $x=1$.
⚠️ Crossing an asymptote
A graph can meet a vertical asymptote $x=a$ at most once (at the point $(a,f(a))$, if $f(a)$ is defined) — and it must run off to $\pm\infty$ from at least one side. But a function may cross a horizontal asymptote, even infinitely many times: $f(x)=\dfrac{\cos x}{x}+1$ oscillates around $y=1$ with ever-decreasing amplitude and crosses it infinitely often (last option in Interactive 1). The algebraic limit laws and the squeeze theorem apply to limits at infinity too.
✏️ Example 1 — Computing limits at infinity
Evaluate $\displaystyle\lim_{x\to\infty}f(x)$ and $\displaystyle\lim_{x\to-\infty}f(x)$, and determine the horizontal and vertical asymptotes of $f(x)=5-\dfrac{2}{x^2}$.
💡 Show solution▶
Limit laws $$\lim_{x\to\infty}\left(5-\frac{2}{x^2}\right)=\lim_{x\to\infty}5-2\left(\lim_{x\to\infty}\frac1x\right)\left(\lim_{x\to\infty}\frac1x\right)=5-2\cdot0\cdot0=5.$$ Similarly $\displaystyle\lim_{x\to-\infty}f(x)=5$.
Horizontal $f$ has the horizontal asymptote $y=5$, approached as $x\to\pm\infty$.
Vertical As $x\to0$, $\dfrac{2}{x^2}\to+\infty$, so $f(x)\to-\infty$ from both sides: $x=0$ is a vertical asymptote.
📘 Definition — Infinite limit at infinity
We say $f$ has an infinite limit at infinity and write $\displaystyle\lim_{x\to\infty}f(x)=\infty$ if $f(x)$ becomes arbitrarily large for $x$ sufficiently large. We write $\displaystyle\lim_{x\to\infty}f(x)=-\infty$ if $f(x)<0$ and $|f(x)|$ becomes arbitrarily large for $x$ sufficiently large. Infinite limits as $x\to-\infty$ are defined similarly.
✏️ Example 2 — Rational functions at $\pm\infty$
For each function, determine the limits as $x\to\infty$ and $x\to-\infty$, and the horizontal asymptote(s).
- 1 $f(x)=\dfrac{3x-1}{2x+5}$ (degrees equal)
- 2 $f(x)=\dfrac{3x^2+2x}{4x^3-5x+7}$ (numerator degree smaller)
- 3 $f(x)=\dfrac{3x^2+4x}{x+2}$ (numerator degree larger)
💡 Show solution▶
(a) Divide numerator and denominator by $x$: $$\lim_{x\to\pm\infty}\frac{3x-1}{2x+5}=\lim_{x\to\pm\infty}\frac{3-1/x}{2+5/x}=\frac{3-0}{2+0}=\frac32.$$ So $y=\tfrac32$ is a horizontal asymptote.
(b) Divide by $x^3$: $$\lim_{x\to\pm\infty}\frac{3x^2+2x}{4x^3-5x+7}=\lim_{x\to\pm\infty}\frac{3/x+2/x^2}{4-5/x^2+7/x^3}=\frac{0+0}{4-0+0}=0.$$ So $y=0$ is a horizontal asymptote.
(c) Divide by $x$: $\displaystyle\lim_{x\to\pm\infty}\frac{3x^2+4x}{x+2}=\lim_{x\to\pm\infty}\frac{3x+4}{1+2/x}$. The denominator $\to1$ while the numerator $\to+\infty$ (as $x\to\infty$) or $-\infty$ (as $x\to-\infty$). Hence $\displaystyle\lim_{x\to\infty}f(x)=\infty$, $\displaystyle\lim_{x\to-\infty}f(x)=-\infty$, and $f$ has no horizontal asymptote.
🧭 Summary — rational functions $p(x)/q(x)$ as $x\to\pm\infty$
- deg $p$ < deg $q$: the limit is $0$; horizontal asymptote $y=0$.
- deg $p$ = deg $q$: the limit is the ratio of leading coefficients; horizontal asymptote $y=\dfrac{a_n}{b_n}$.
- deg $p$ > deg $q$: the limit is $\pm\infty$; no horizontal asymptote.
- Method: divide numerator and denominator by the highest power of $x$ in the denominator, then use $\displaystyle\lim_{x\to\pm\infty}\frac{1}{x^k}=0$.
- Vertical asymptotes: zeros of the denominator (after cancelling common factors).
Exercises
I
Evaluate the limit.
- 1 $\displaystyle\lim_{x\to\infty}\frac{1}{3x+6}$
- 2 $\displaystyle\lim_{x\to\infty}\frac{2x-5}{4x}$
- 3 $\displaystyle\lim_{x\to\infty}\frac{x^2-2x+5}{x+2}$
- 4 $\displaystyle\lim_{x\to-\infty}\frac{3x^3-2x}{x^2+2x+8}$
- 5 $\displaystyle\lim_{x\to-\infty}\frac{x^4-4x^3+1}{2-2x^2-7x^4}$
- 6 $\displaystyle\lim_{x\to\infty}\frac{3x}{\sqrt{x^2+1}}$
- 7 $\displaystyle\lim_{x\to-\infty}\frac{\sqrt{4x^2-1}}{x+2}$
- 8 $\displaystyle\lim_{x\to\infty}\frac{4x}{\sqrt{x^2-1}}$
Numerical check: a table of values heading to $\pm\infty$
💡 Show solutions▶
1 Divide by $x$: $\dfrac{1/x}{3+6/x}\to\dfrac{0}{3}=0$.
2 $\dfrac{2x-5}{4x}=\dfrac{2-5/x}{4}\to\dfrac24=\dfrac12$.
3 Divide by $x$: $\dfrac{x-2+5/x}{1+2/x}$; the numerator $\to\infty$, the denominator $\to1$, so the limit is $\infty$.
4 Divide by $x^2$: $\dfrac{3x-2/x}{1+2/x+8/x^2}$; as $x\to-\infty$ the numerator $\to-\infty$ and the denominator $\to1$, so the limit is $-\infty$.
5 Degrees equal (4): divide by $x^4$: $\dfrac{1-4/x+1/x^4}{2/x^4-2/x^2-7}\to\dfrac{1}{-7}=-\dfrac17$.
6 For $x>0$, $\sqrt{x^2+1}=x\sqrt{1+1/x^2}$, so $\dfrac{3x}{x\sqrt{1+1/x^2}}=\dfrac{3}{\sqrt{1+1/x^2}}\to3$.
7 For $x<0$, $\sqrt{x^2}=|x|=-x$, so $\sqrt{4x^2-1}=-x\sqrt{4-1/x^2}$. Then $\dfrac{-x\sqrt{4-1/x^2}}{x(1+2/x)}=\dfrac{-\sqrt{4-1/x^2}}{1+2/x}\to-\dfrac{\sqrt4}{1}=-2$. (The sign flip from $|x|=-x$ is the key step.)
8 For $x>1$, $\sqrt{x^2-1}=x\sqrt{1-1/x^2}$, so $\dfrac{4}{\sqrt{1-1/x^2}}\to4$.
II
Examine the graphs. Identify where the vertical asymptotes are located.
The five graphs
💡 Show solutions▶
(1) The graph shoots to $+\infty$ just right of $x=1$ and to $-\infty$ just left of it: vertical asymptote $x=1$. (Graph of $y=\dfrac{1}{x-1}$.)
(2) Vertical asymptote $x=-2$. (Graph of $y=\dfrac{x}{x+2}$; it also levels off at $y=1$.)
(3) Two vertical asymptotes, $x=-1$ and $x=1$. (Graph of $y=\dfrac{1}{x^2-1}$.)
(4) Vertical asymptote $x=3$; both sides go to $+\infty$. (Graph of $y=\dfrac{1}{(x-3)^2}$.)
(5) Vertical asymptote $x=0$; both sides go to $+\infty$ near $0$. (Graph of $y=\dfrac{x+1}{x^2}$, which crosses its horizontal asymptote $y=0$ at $x=-1$.)
III
Find the horizontal and vertical asymptotes.
- 1 $f(x)=x-\dfrac9x$
- 2 $f(x)=\dfrac{1}{1-x^2}$
- 3 $f(x)=\dfrac{x^3}{4-x^2}$
- 4 $f(x)=\dfrac{x^2+3}{x^2+1}$
- 5 $f(x)=\dfrac{1}{x^3+x^2}$
Asymptote Finder — a point slides toward each vertical asymptote from both sides
$x$—
$f(x)$—
vertical—
horizontal—
💡 Show solutions▶
1 $f(x)=\dfrac{x^2-9}{x}$. Vertical: $x=0$ ($\lim_{x\to0^+}f=-\infty$, $\lim_{x\to0^-}f=+\infty$). Horizontal: none — $\lim_{x\to\pm\infty}f(x)=\pm\infty$ (the graph follows the slant line $y=x$).
2 $1-x^2=0$ at $x=\pm1$: vertical $x=-1$ and $x=1$. Degree of numerator $<$ denominator: horizontal $y=0$.
3 $4-x^2=0$ at $x=\pm2$: vertical $x=-2$ and $x=2$. Numerator degree $3>2$: $\lim_{x\to\infty}f=-\infty$, $\lim_{x\to-\infty}f=+\infty$, so no horizontal asymptote.
4 $x^2+1>0$ for all $x$: no vertical asymptote. Equal degrees, leading coefficients $1/1$: horizontal $y=1$.
5 $x^3+x^2=x^2(x+1)=0$ at $x=0,-1$: vertical $x=0$ and $x=-1$. Numerator degree $0<3$: horizontal $y=0$.
IV
Evaluate each limit. For the radicals, remember that $\sqrt{x^2}=|x|$, which equals $-x$ when $x<0$.
- 1 $\displaystyle\lim_{x\to\infty}\frac{5x^2-3x+1}{2x^2+7}$
- 2 $\displaystyle\lim_{x\to-\infty}\frac{4x-x^3}{2x^3+x^2-1}$
- 3 $\displaystyle\lim_{x\to\infty}\frac{x+4}{x^2-3}$
- 4 $\displaystyle\lim_{x\to-\infty}\frac{2x^2+1}{x-3}$
- 5 $\displaystyle\lim_{x\to\infty}\frac{\sqrt{x^2+1}}{x}$
- 6 $\displaystyle\lim_{x\to-\infty}\frac{\sqrt{x^2+1}}{x}$
- 7 $\displaystyle\lim_{x\to-\infty}\frac{x+\sqrt{9x^2+2}}{2x-1}$ (and compare with $x\to\infty$)
- 8 $\displaystyle\lim_{x\to\infty}\left(\sqrt{x^2+3x}-x\right)$
- 9 Describe the end behaviour of $p(x)=-2x^3+5x^2-1$.
Numerical check: a table of values heading to $\pm\infty$
💡 Show solutions▶
1 Equal degrees (2): divide by $x^2$: $\dfrac{5-3/x+1/x^2}{2+7/x^2}\to\dfrac52$. The limit is the ratio of leading coefficients.
2 Equal degrees (3): divide by $x^3$: $\dfrac{4/x^2-1}{2+1/x-1/x^3}\to\dfrac{-1}{2}=-\dfrac12$. The direction $-\infty$ does not matter here, because every leftover term still goes to $0$.
3 Numerator degree $1<2$: divide by $x^2$: $\dfrac{1/x+4/x^2}{1-3/x^2}\to\dfrac01=0$.
4 Numerator degree $2>1$: divide by $x$: $\dfrac{2x+1/x}{1-3/x}$. The numerator behaves like $2x\to-\infty$ and the denominator $\to1$, so the limit is $-\infty$.
5 For $x>0$, $\sqrt{x^2+1}=|x|\sqrt{1+1/x^2}=x\sqrt{1+1/x^2}$, so $\dfrac{\sqrt{x^2+1}}{x}=\sqrt{1+1/x^2}\to1$.
6 For $x<0$, $|x|=-x$, so $\sqrt{x^2+1}=-x\sqrt{1+1/x^2}$ and $\dfrac{\sqrt{x^2+1}}{x}=-\sqrt{1+1/x^2}\to-1$. Same formula, opposite answer: the function has two horizontal asymptotes, $y=1$ and $y=-1$.
7 For $x<0$, $\sqrt{9x^2+2}=-x\sqrt{9+2/x^2}$. Dividing top and bottom by $x$:
$$\frac{1-\sqrt{9+2/x^2}}{2-1/x}\to\frac{1-3}{2}=-1.$$
For $x>0$ the sign is $+$: $\dfrac{1+\sqrt{9+2/x^2}}{2-1/x}\to\dfrac{1+3}{2}=2$. Two different horizontal asymptotes: $y=-1$ on the left, $y=2$ on the right.
8 This is $\infty-\infty$, so multiply by the conjugate:
$$\sqrt{x^2+3x}-x=\frac{(x^2+3x)-x^2}{\sqrt{x^2+3x}+x}=\frac{3x}{x\sqrt{1+3/x}+x}=\frac{3}{\sqrt{1+3/x}+1}\to\frac32.$$
9 The leading term $-2x^3$ controls the end behaviour. As $x\to\infty$, $-2x^3\to-\infty$; as $x\to-\infty$, $x^3\to-\infty$ so $-2x^3\to+\infty$. Hence $\displaystyle\lim_{x\to\infty}p(x)=-\infty$ and $\displaystyle\lim_{x\to-\infty}p(x)=+\infty$: the graph rises on the left and falls on the right.
V
Find all horizontal and vertical asymptotes, and give the one-sided infinite limits at each vertical asymptote.
- 1 $f(x)=\dfrac{2x^2}{x^2-4}$
- 2 $g(x)=\dfrac{x-1}{x^2-1}$ (careful: one zero of the denominator is not an asymptote)
- 3 $h(x)=\dfrac{x+2}{x-3}$
- 4 $q(x)=\dfrac{3x-2}{\sqrt{x^2+1}}$
- 5 Evaluate $\displaystyle\lim_{x\to0}\frac{x-1}{x^2}$.
Asymptote Finder — a point slides toward each vertical asymptote from both sides
$x$—
$f(x)$—
vertical—
horizontal—
💡 Show solutions▶
1 Horizontal: equal degrees, ratio $\tfrac21$, so $y=2$ at both ends. Vertical: $x^2-4=0$ at $x=\pm2$, and the numerator $2x^2=8\neq0$ there. Near $x=2$: $x^2-4\to0^+$ from the right and $0^-$ from the left, so $\displaystyle\lim_{x\to2^+}f=+\infty$, $\displaystyle\lim_{x\to2^-}f=-\infty$. Near $x=-2$: $\displaystyle\lim_{x\to-2^-}f=+\infty$, $\displaystyle\lim_{x\to-2^+}f=-\infty$.
2 Factor: $g(x)=\dfrac{x-1}{(x-1)(x+1)}=\dfrac{1}{x+1}$ for $x\neq1$. At $x=1$ the factor cancels, so $\displaystyle\lim_{x\to1}g=\tfrac12$: a hole at $\left(1,\tfrac12\right)$, not an asymptote. At $x=-1$: $\displaystyle\lim_{x\to-1^+}g=+\infty$, $\displaystyle\lim_{x\to-1^-}g=-\infty$, so vertical $x=-1$. Degree $1<2$: horizontal $y=0$.
3 Vertical $x=3$: the numerator is $5>0$ there, so $\displaystyle\lim_{x\to3^+}h=+\infty$ and $\displaystyle\lim_{x\to3^-}h=-\infty$. Horizontal $y=1$ (equal degrees, ratio $\tfrac11$).
4 $\sqrt{x^2+1}>0$ always: no vertical asymptote. Dividing by $|x|$: for $x>0$, $\dfrac{3-2/x}{\sqrt{1+1/x^2}}\to3$; for $x<0$, $\sqrt{x^2+1}=-x\sqrt{1+1/x^2}$ gives $\dfrac{3-2/x}{-\sqrt{1+1/x^2}}\to-3$. Two horizontal asymptotes: $y=3$ (right) and $y=-3$ (left).
5 As $x\to0$ the numerator $\to-1$ while $x^2\to0^+$ from both sides. A negative number divided by a tiny positive number is hugely negative, so $\displaystyle\lim_{x\to0}\frac{x-1}{x^2}=-\infty$ (the same from each side, so the two-sided infinite limit exists).
VI
Models that settle down. Find each long-run limit and explain what it means.
- 1 Drug concentration. The concentration of a drug in the blood $t$ hours after an injection is $C(t)=\dfrac{8t}{t^2+1}$ mg/L. Find $\displaystyle\lim_{t\to\infty}C(t)$. (The peak is $C(1)=4$ mg/L; compute $C(10)$ to see the decline.)
- 2 Learning curve. A trainee's typing speed after $t$ minutes of practice is $S(t)=\dfrac{30t}{200+t}$ words per minute. Find $\displaystyle\lim_{t\to\infty}S(t)$ and the time needed to reach half of it.
- 3 Population. A fish population in a reservoir is modelled by $P(t)=\dfrac{5000}{1+49e^{-0.3t}}$, $t$ in months. Find $P(0)$ and $\displaystyle\lim_{t\to\infty}P(t)$.
- 4 Average cost. A workshop has fixed costs of $500$ OMR and each item costs $2$ OMR to make, so the average cost per item for $x$ items is $A(x)=\dfrac{500+2x}{x}$. Find $\displaystyle\lim_{x\to\infty}A(x)$.
- 5 Conceptual. True or false: “a graph can never cross its horizontal asymptote.” Give a reason or a counterexample.
Watch each model level off
input—
output—
vertical—
long-run level—
💡 Show solutions▶
1 Degree $1<2$: $\dfrac{8t}{t^2+1}=\dfrac{8/t}{1+1/t^2}\to0$. So $\displaystyle\lim_{t\to\infty}C(t)=0$: the drug is eventually cleared from the blood. Already $C(10)=\dfrac{80}{101}\approx0.79$ mg/L, about a fifth of the peak.
2 Divide by $t$: $\dfrac{30}{200/t+1}\to30$. The trainee's speed approaches $30$ words per minute but never reaches it. Half of it: $\dfrac{30t}{200+t}=15\Rightarrow30t=3000+15t\Rightarrow t=200$ minutes.
3 $P(0)=\dfrac{5000}{1+49}=100$ fish. As $t\to\infty$, $e^{-0.3t}\to0$, so $P(t)\to\dfrac{5000}{1+0}=5000$. The reservoir's carrying capacity is $5000$ fish: $y=5000$ is a horizontal asymptote.
4 $A(x)=\dfrac{500}{x}+2\to0+2=2$. As production grows, the fixed cost is spread over more items and the average cost falls toward the $2$ OMR cost of making one item — never below it.
5 False. The asymptote only describes behaviour as $x\to\pm\infty$; in the middle the graph can do anything. For example $y=\dfrac{x+1}{x^2}$ has horizontal asymptote $y=0$ (degree $1<2$) but crosses it at $x=-1$. (Interactive 1's $\tfrac{\cos x}{x}+1$ even crosses $y=1$ infinitely often.)
🔬 Research–Teaching NexusResearch-based · optional, about 30 minutes
Research question. Can you predict the room temperature by watching a cup of karak tea cool?
Why it matters. Newton's law of cooling, $T(t)=T_a+(T_0-T_a)e^{-kt}$, is used in food safety, forensic science and electronics cooling; the room temperature $T_a$ is the horizontal asymptote.
What to do.
- Pour hot tea or coffee into a cup and measure its temperature every 2 minutes for about 25 minutes (a kitchen thermometer works).
- Enter the data in Desmos or a spreadsheet and plot temperature against time.
- Fit $T(t)=T_a+(T_0-T_a)e^{-kt}$ (Desmos regression: y1 ~ a + b e^{-k x1}). Read off $\lim_{t\to\infty}T(t)=T_a$.
- Compare your $T_a$ with a thermometer reading of the room.
Tools: Kitchen thermometer, phone timer, Desmos
Hand in: A half-page note: your data, the fitted model, the predicted and measured room temperature, and one limitation.
Going further. Anywhere: repeat with a lid on the cup and compare the values of $k$.
Instructor note: Students should notice the curve never reaches $T_a$ in finite time; the asymptote is estimated, and short data sets give less reliable $T_a$.
🧪 Self-check quiz
0/5 correct
Q1. $\displaystyle\lim_{x\to\infty}\frac{5x^2-1}{2x^2+x}=$
Q2. Which function has no horizontal asymptote?
Q3. Can a graph cross its horizontal asymptote?
Q4. The vertical asymptotes of $\dfrac{1}{x^3+x^2}$ are
Q5. $\displaystyle\lim_{x\to-\infty}\frac{\sqrt{4x^2-1}}{x+2}=$