Single-Variable-Calculus · Unit 2 · Video 5 · Interactive Practice

Infinity Has a Sign: The Hyperbola, Its Derivative, and a Limit That Does Not Exist

IKey Formulas

Statement	Name	What it says
$\lim\limits_{x \to 0^{+}} \dfrac{1}{x} = +\infty \qquad \lim\limits_{x \to 0^{-}} \dfrac{1}{x} = -\infty$	Infinite discontinuity	The two sides run in opposite directions
$\dfrac{d}{dx}\left(\dfrac{1}{x}\right) = -\dfrac{1}{x^{2}}$	Derivative of the hyperbola	Negative for every $x \neq 0$
$\lim\limits_{x \to 0}\left(-\dfrac{1}{x^{2}}\right) = -\infty$	A legitimate two-sided limit	Both sides agree, so no side need be named
$f(-x) = -f(x) \;\Rightarrow\; f'(-x) = f'(x)$	Odd $\Rightarrow$ even	$x^{-1}$ is an odd power, $x^{-2}$ an even one

Key Insight: Infinity carries a sign. Writing $\lim\limits_{x \to 0} \dfrac{1}{x} = \infty$ drops both the side and the sign — and because the two sides disagree, no value, finite or infinite, describes that two-sided limit.

IIVisualization 1 — Which Way Does $1/x$ Go?

Both branches run off to infinity at $x = 0$ , but not in the same direction.

Side x 3

IIIVisualization 2 — The Derivative Graph Records Slope

The height of the $f'$ graph at $x$ is the slope of the $f$ graph at the same $x$ .

Base point x₀ 1.00

💡 $1/x$ is odd and $-1/x^{2}$ is even: differentiating an odd function always produces an even one, matching the odd power $x^{-1}$ and the even power $x^{-2}$ .

IVVisualization 3 — When Is a Two-Sided Limit Legitimate?

A two-sided limit statement is legitimate only when both one-sided limits agree.

Function Step 0

💡 $\sin\dfrac{1}{x}$ is the fourth kind of discontinuity — the ugly ones — and the only kind with no one-sided limit at all, not even an infinite one.

VQuiz Questions

Problem 1 · One Side of the Hyperbola

Given: $f(x) = \dfrac{1}{x}$ — evaluate $\lim\limits_{x \to 0^{-}} \dfrac{1}{x}$ .

✅ Correct! At

x = -0.1, -0.01, -0.001

the values are

-10, -100, -1000

: they drop without bound.

❌ Wrong side.

+\infty

is the limit from the right. The notation

x \to 0^{-}

means

x

is negative, and

1/x

is negative there.

❌ Not quite.

1/x

is near

0

when

x

is large. Dividing by a tiny number makes the quotient huge, not small.

❌ Too pessimistic. The two-sided limit does not exist, but this one-sided limit heads in a very definite direction — say which.

❌ Not quite. Test

x = -0.001

1/x = -1000

Show solution

Approach $0$ through negative values:

\frac{1}{-0.1} = -10, \qquad \frac{1}{-0.01} = -100, \qquad \frac{1}{-0.001} = -1000

The numerator is fixed at $1$ and the denominator is negative and shrinking toward $0$ , so the quotient is negative and grows without bound in size:

\lim_{x \to 0^{-}} \frac{1}{x} = -\infty

This is the left branch of the hyperbola, which plunges down the negative side of the $y$ -axis.

Problem 2 · The Statement That Drops the Sign

Given: a student writes $\lim\limits_{x \to 0} \dfrac{1}{x} = \infty$ . Which criticism is the correct one?

✅ Correct! The two one-sided limits disagree, so no single value — finite or infinite — describes the two-sided limit.

❌ Not harmless here.

+\infty

is an infinite number of dollars and

-\infty

an infinite amount of debt; the hyperbola does both at once, one on each side.

❌ A sign alone will not save it.

+\infty

is only what the right branch does. The left branch drops to

-\infty

, so no sign makes the two-sided statement true.

❌ Close, but incomplete. Naming the side fixes half of it — the

\infty

still needs its

+

, and the original claim was about both sides, where nothing true can be said.

❌ Not quite. Compare the two sides separately before deciding what the two-sided statement can say.

Show solution

Compute each side on its own:

\lim_{x \to 0^{+}} \frac{1}{x} = +\infty, \qquad \lim_{x \to 0^{-}} \frac{1}{x} = -\infty

A two-sided limit exists only when both one-sided limits agree. Here they are as far apart as two directions can be, so the two-sided limit does not exist — and $\infty$ with no sign and no side hides exactly the disagreement that matters.

The two correct statements are the one-sided ones. Whenever a limit heads in a definite direction, say which direction.

Problem 3 · Where the Derivative Goes

Given: $f(x) = \dfrac{1}{x}$ , so $f'(x) = -\dfrac{1}{x^{2}}$ .

What is $\lim\limits_{x \to 0^{+}} f'(x)$ ?

Is $\lim\limits_{x \to 0} f'(x) = -\infty$ a legitimate statement?

✅ Correct!

x^{2} > 0

for every

x \neq 0

, so

-1/x^{2}

is negative on both sides and plunges to

-\infty

on both — the two-sided statement is earned.

❌ Check the sign. As

x \to 0^{+}

x^{2}

shrinks to

0

, so

1/x^{2}

grows without bound — and the minus sign in front sends it downward.

❌ Check both sides. Squaring destroys the sign of

x

, so

-1/x^{2}

behaves identically for

x < 0

and

x > 0

Show solution

From the right: at $x = 0.1, 0.01, 0.001$ ,

-\frac{1}{x^{2}} = -100, \; -10\,000, \; -1\,000\,000 \;\longrightarrow\; -\infty

From the left: at $x = -0.1, -0.01, -0.001$ the squares are the same numbers, so the values are the same: also $-\infty$ .

Because the two one-sided limits agree, the two-sided statement is legitimate:

\lim_{x \to 0}\left(-\frac{1}{x^{2}}\right) = -\infty

Contrast this with $1/x$ , where the sides disagreed and no two-sided statement was available. Note also that this is the graph of a derivative: $f$ splits up and down at $0$ , while $f'$ plunges downward on both sides — the derivative graph looks nothing like the function's.

Problem 4 · Sorting the Zoo

Given: $g(x) = \dfrac{1}{x^{3}}$ , whose derivative is $g'(x) = -\dfrac{3}{x^{4}}$ , together with $\sin\dfrac{1}{x}$ — all three examined at $x = 0$ .

Which one has a legitimate two-sided infinite limit at $x = 0$ ?

Which one has no one-sided limit at all at $x = 0$ ?

✅ Correct! The even power

x^{4}

makes

g'

behave the same on both sides, the odd power

x^{3}

makes

g

split, and

\sin(1/x)

never settles down at all.

❌ Check the parity of the power. An odd power keeps the sign of

x

, so its two sides disagree; only an even power in the denominator makes both sides do the same thing.

❌ Not that one. Running off to

-\infty

+\infty

is a definite direction. Look for the function whose values keep swinging instead of heading anywhere.

Show solution

$g(x) = 1/x^{3}$ : $x^{3}$ keeps the sign of $x$ , so

\lim_{x \to 0^{+}} \frac{1}{x^{3}} = +\infty, \qquad \lim_{x \to 0^{-}} \frac{1}{x^{3}} = -\infty

Both one-sided limits exist, but they disagree — no two-sided statement, just like $1/x$ .

$g'(x) = -3/x^{4}$ : $x^{4} > 0$ for every $x \neq 0$ , so the values are negative and huge on both sides:

\lim_{x \to 0}\left(-\frac{3}{x^{4}}\right) = -\infty

Both sides agree, so the two-sided statement is legitimate. (As expected: $g$ is odd, so $g'$ is even.)

$\sin(1/x)$ : as $x \to 0$ the input $1/x$ races through every angle, so the values oscillate through $[-1, 1]$ infinitely often. They never approach a number, and they never run off to $\pm\infty$ either — neither one-sided limit exists.

Solved: 0 / 4

Infinity Has a Sign: The Hyperbola, Its Derivative, and a Limit That Does Not Exist

IKey Formulas

IIVisualization 1 — Which Way Does 1/x1/x1/x Go?

IIIVisualization 2 — The Derivative Graph Records Slope

IVVisualization 3 — When Is a Two-Sided Limit Legitimate?

VQuiz Questions

IIVisualization 1 — Which Way Does $1/x$ Go?