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What happens far out, what happens beside a pole, and why the two questions are told apart.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to evaluate a limit at infinity by dividing through by the highest power in the denominator, read a horizontal asymptote off the degrees, handle a square root by pulling out $|x|$ and checking the two ends separately, decide the sign of an infinite limit on each side of a pole, tell a pole from a removable hole by whether the factor cancels, and find a slant asymptote by division. You will also be able to say why $\lim = \infty$ is not a limit existing, and what that costs you.
A limit at infinity asks what the outputs approach as the inputs run away: $\lim_{x \to \infty} f(x) = L$ means $f(x)$ can be made as close to $L$ as you like by taking $x$ large enough. The quantifiers are the definition you already know with "close enough to $a$" replaced by "far enough out": for every $\varepsilon > 0$ there is an $M$ with $x > M \implies |f(x) - L| < \varepsilon$. When such an $L$ exists, $y = L$ is a horizontal asymptote.
An infinite limit asks the opposite: $\lim_{x \to a} f(x) = \infty$ means $f(x)$ exceeds any bound you name once $x$ is close enough to $a$. This is not a limit existing — $\infty$ is not a real number — it is a precise description of how the limit fails, and $x = a$ is a vertical asymptote.
The two are easy to confuse because both use the same symbol, and they answer opposite questions: one is about large $x$, the other about large $y$. A function may have either, both or neither.
Rational functions. For $\dfrac{p(x)}{q(x)}$ as $x \to \pm\infty$:
| Degrees | Limit | Asymptote |
|---|---|---|
| $\deg p < \deg q$ | $0$ | $y = 0$ |
| $\deg p = \deg q$ | ratio of leading coefficients | $y = $ that ratio |
| $\deg p = \deg q + 1$ | none | a slant asymptote, found by division |
| $\deg p > \deg q + 1$ | none | none of the above |
The technique behind every row is the same: divide every term by the highest power in the denominator, and read off what survives.
Another way: picture
A horizontal asymptote is a level the graph settles towards as it travels right or left; a vertical asymptote is a wall it climbs beside as $x$ approaches a point. Two different directions of travel, two different questions. And the graph is allowed to cross a horizontal asymptote — repeatedly, even — because approaching a level says nothing about staying on one side of it.
Another way: steps
For a limit at infinity:
For an infinite limit at a point:
$\sqrt{x^2} = |x|$, not $x$. This is the single most common error in limits at infinity, and it is invisible when $x \to +\infty$ because there $|x| = x$ and the wrong step gives the right answer.
It bites at the other end. For $f(x) = \dfrac{3x}{\sqrt{4x^2 + 1}}$:
Two horizontal asymptotes on one graph, $y = \tfrac32$ and $y = -\tfrac32$, and the difference is entirely the sign that the absolute value supplies. A function that is a ratio of polynomials cannot do this; one with a root can, and checking both ends separately is the habit that catches it.
Treating $\infty/\infty$ as $1$, or $\infty - \infty$ as $0$. Both are indeterminate, for the same reason $0/0$ is: the answer depends on relative rates, which the form does not record. $\dfrac{x^2}{x} \to \infty$, $\dfrac{x}{x^2} \to 0$, $\dfrac{2x}{x} \to 2$ — all $\infty/\infty$.
Reading $\frac{c}{0}$ as indeterminate. With $c \ne 0$ it is not. The size is settled and only the sign needs a one-sided look.
Assuming a graph cannot cross a horizontal asymptote. $\dfrac{\sin x}{x}$ crosses $y = 0$ infinitely often and still tends to $0$. Asymptote means approaches, not is bounded by.
Dropping the absolute value under a root. Right at $+\infty$, wrong at $-\infty$, and the error hides.
Cancelling to find vertical asymptotes. $\dfrac{(x-1)(x+2)}{(x-1)(x-3)}$ has a vertical asymptote at $x = 3$ only. At $x = 1$ the factor cancels and the break is a removable hole — a missing point, not a wall.
Writing $\lim_{x \to 0} \dfrac{1}{x^2} = \infty$ is standard and useful, and it does not mean the limit exists. $\infty$ is not a real number, and the equation is shorthand for a precise statement about the limit failing in a particular way: the values exceed every bound.
The distinction has teeth. "The limit exists" permits the limit laws; "the limit is infinite" permits none of them, which is why $\infty - \infty$ and $0 \times \infty$ cannot be evaluated by arithmetic. And a function can fail to have a limit without being infinite at all — $\sin(1/x)$ near $0$ is bounded and has no limit — so "the limit is infinite" is more informative than "no limit", and neither is "the limit exists".
$\lim_{x \to \infty} \left(\sqrt{x^2 + x} - x\right)$ is of the form $\infty - \infty$, so nothing can be read off.
Indeterminate: do algebra.
Multiply by the conjugate over itself: $\dfrac{(x^2 + x) - x^2}{\sqrt{x^2+x} + x} = \dfrac{x}{\sqrt{x^2+x}+x}$.
The conjugate turns a difference into a quotient.
Divide by $x$ (positive here): $\dfrac{1}{\sqrt{1 + 1/x} + 1} \to \dfrac{1}{2}$. So the two roots pull apart by half in the limit, not by nothing and not by infinity.
$f(x) = \dfrac{x^2 - 1}{x^2 - 4}$. Far out, equal degrees give $y = 1$: a horizontal asymptote, approached from above on one side and below on the other.
Large $x$.
At $x = 2$ the denominator vanishes and the numerator does not ($3 \ne 0$), so there is a vertical asymptote, with the sign flipping across it.
Large $y$.
At $x = -2$ the same, since $(-2)^2 - 1 = 3 \ne 0$. Nothing cancels, so both zeros of the denominator are walls rather than holes.
Check cancelling before calling it an asymptote.
Equal degrees, so divide every term by $x^3$.
The standard first move.
$\dfrac{5 - 1/x^2}{2 + 7/x}$, and every term with an $x$ underneath vanishes.
The limit is $\tfrac52$ — the ratio of the leading coefficients, as the rule promises.
Match each degree comparison to the rational function's end behavior.
| Limit is the ratio of leading coefficients | Limit is 0 | A slant asymptote may result after division | No horizontal or slant asymptote from a linear quotient | |
|---|---|---|---|---|
| Equal numerator and denominator degrees | ||||
| Numerator degree is lower | ||||
| Numerator degree is one higher | ||||
| Numerator degree exceeds denominator by at least two |
Match each behaviour to its asymptote conclusion.
| Vertical asymptote $x=2$ | Horizontal asymptote $y=3$ | The rational-function limit is 0 | Magnitude grows without bound (or has no finite limit) | |
|---|---|---|---|---|
| $f(x)\to\infty$ as $x\to2$ | ||||
| $f(x)\to3$ as $x\to\infty$ | ||||
| degree numerator < degree denominator | ||||
| degree numerator > degree denominator |
Match each limiting behavior to the asymptote conclusion it supports.
| $y=3$ is a right-hand horizontal asymptote | $x=2$ is a vertical asymptote | $y=2x-1$ is a slant asymptote | The point may be a removable hole rather than a vertical asymptote | |
|---|---|---|---|---|
| $\lim_{x\to\infty}f(x)=3$ | ||||
| $\lim_{x\to2^-}f(x)=-\infty$ | ||||
| $f(x)-(2x-1)\to0$ as $x\to\infty$ | ||||
| A denominator zero cancels with the numerator |
Find $\displaystyle\lim_{x \to \infty} \frac{9x^2 + 2x}{5x^2 - 2}$.
Answer:
Find $\displaystyle\lim_{x \to \infty} \frac{8x + 1}{8x^2 + 5}$, or give $0$ if it is zero.
Answer:
$f(x) = \dfrac{6x^2 + 2}{x}$. Far out, $f$ approaches a line $y = ax + b$. What is $a$?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
For $f(x) = \dfrac{3x^2 + 1}{x^2 - 4}$, what is the horizontal asymptote as $x \to \infty$?
You can find horizontal, vertical and slant asymptotes and say which question each answers. Say in your own words why $\sqrt{x^2}$ is $|x|$ and where that changes an answer. Next: the derivative, defined as the limit this unit has been building towards.
8. Your turn: $\lim_{x \to \infty} \dfrac{5x^3 - x}{2x^3 + 7x^2}$, step 3