What's the difference between \lim_{x \to 3} f(x) = \infty and f(3) = \infty?

The limit describes behavior as x approaches 3, not what happens exactly at 3. Since infinity is not a real number, f(3) = \infty is undefined—the function typically has no output at x = 3. However, \lim_{x \to 3} f(x) can still equal \infty. For instance, f(x) = \frac{1}{x-3} is undefined at x = 3, yet the one-sided limits are infinite.

Can one side of a limit be infinite while the other is finite?

No; if either one-sided limit is infinite, the two-sided limit does not exist. However, you can have one one-sided limit be +\infty and the other be -\infty, as in \lim_{x \to 0^{-}} \frac{1}{x} = -\infty and \lim_{x \to 0^{+}} \frac{1}{x} = +\infty. In this case, the two-sided limit \lim_{x \to 0} \frac{1}{x} is undefined.

How do I know if a limit is +\infty or -\infty just by looking at the expression?

Factor the numerator and denominator, then determine the sign of each piece as x approaches the target value from each direction. The product of the numerator's sign and the denominator's sign tells you whether the result is +\infty or -\infty. Creating a sign chart or testing points slightly to the left and right of the target value helps determine signs accurately.

Are infinite limits the same as discontinuities?

Infinite limits indicate a vertical asymptote, which is one specific type of discontinuity, but not all discontinuities are infinite. A jump discontinuity (where left and right limits are finite but different) or a removable discontinuity (a hole in the graph) do not involve infinite limits. Functions with infinite limits are always discontinuous, but discontinuity does not always mean an infinite limit exists.

Infinite Limits

Infinite Limits describe when a function grows unboundedly as the input approaches a specific value, letting you identify and analyze vertical asymptotes.

\lim_{x \to a} f(x) = \pm\infty

Solve a problem with Infinite Limits

Type the problem. The solver will use Infinite Limits where Infinite Limits is the right tool, and tell you when it is not.

What each symbol means

What Infinite Limits takes

Symbol	Meaning
$f$	The function being analyzed—this is the rule or formula you're studying, not a single number; forgetting this distinction leads to applying the limit machinery to constants instead of expressions.
$x$	The independent variable (input); in a limit, $x$ approaches $a$ arbitrarily closely but never actually equals $a$ , and confusing this causes students to think the function must be defined exactly at $a$ for the limit to exist.
$a$	The specific value the input approaches; the limit describes what happens as $x$ gets close to $a$ , not necessarily what happens at $x = a$ itself, so the function can be undefined at that point even when the limit is infinite.

When to use it

Use this when a function's outputs become unbounded as the input approaches a particular value, indicating a vertical asymptote.

Level

Usually taught in: Calculus I

Worked examples

1. One-sided limits of a simple rational function

Problem

Determine

\lim_{x \to 2} \frac{1}{x-2}

$x - 2 \to 0^{+} \text{ as } x \to 2^{+}$
When $x$ approaches 2 from the right (values slightly greater than 2), the denominator approaches a small positive number.
$\frac{1}{x-2} \to +\infty \text{ as } x \to 2^{+}$
A small positive number in the denominator makes the fraction grow without bound in the positive direction.
$x - 2 \to 0^{-} \text{ as } x \to 2^{-}$
When $x$ approaches 2 from the left (values slightly less than 2), the denominator approaches a small negative number.
$\frac{1}{x-2} \to -\infty \text{ as } x \to 2^{-}$
A small negative number in the denominator makes the fraction grow without bound in the negative direction.

Answer: $\lim_{x \to 2^{+}} \frac{1}{x-2} = +\infty \text{ and } \lim_{x \to 2^{-}} \frac{1}{x-2} = -\infty$

The left and right limits differ (one is $+\infty$ , the other is $-\infty$ ), so there is no single two-sided infinite limit. The function has a vertical asymptote at $x = 2$ , and the sign of infinity depends on which direction you approach from.

2. Squared denominator with positive infinity from both sides

Problem

Evaluate

\lim_{x \to -1} \frac{5}{(x+1)^2}

$(x+1)^2 \to 0^{+} \text{ as } x \to -1$
Since $(x+1)^2$ is always non-negative and equals zero only at $x = -1$ , it approaches 0 from the positive side regardless of approach direction.
$\frac{5}{(x+1)^2} = \frac{\text{positive}}{\text{positive}} \to +\infty$
A positive numerator divided by a small positive denominator yields a positive result growing without bound.
$\lim_{x \to -1^{+}} \frac{5}{(x+1)^2} = +\infty \text{ and } \lim_{x \to -1^{-}} \frac{5}{(x+1)^2} = +\infty$
Both one-sided limits approach $+\infty$ because the denominator is positive from both sides.

Answer: $\lim_{x \to -1} \frac{5}{(x+1)^2} = +\infty$

When the denominator becomes zero but is always positive near the point, both one-sided limits equal $+\infty$ , so we can write the two-sided limit as $+\infty$ . The squaring in the denominator ensures the sign is always positive regardless of approach direction.

3. Garden design: planting density approaching infinity as width shrinks

Problem

A farmer plans to distribute exactly 50 plants in a rectangular garden of width

w

meters. The planting density is

D(w) = \frac{50}{w}

plants per square meter. Determine

\lim_{w \to 0^{+}} \frac{50}{w}

$D(w) = \frac{50}{w} \text{ with } w > 0$
The density function depends on the width; 50 is the fixed total number of plants to accommodate.
$w = 0.1 \implies D(0.1) = \frac{50}{0.1} = 500$
At a width of 0.1 meter, the density must be 500 plants per square meter to fit all 50 plants.
$w = 0.01 \implies D(0.01) = \frac{50}{0.01} = 5000$
At a width of 0.01 meter, the required density becomes 5000 plants per square meter—clearly impractical.
$\text{As } w \to 0^{+}, \text{ the fraction } \frac{50}{w} \text{ grows without bound}$
There is no smallest positive width; we can always make $w$ smaller, forcing the density to become arbitrarily large.

Answer: $\lim_{w \to 0^{+}} \frac{50}{w} = +\infty$

This infinite limit shows that fitting a fixed number of plants into an increasingly narrow garden requires the planting density to grow without bound. The limit identifies the boundary of feasible designs: widths close to zero make the problem physically impossible, illustrating why infinite limits matter in applied engineering.

Common mistakes

Where Infinite Limits usually goes wrong

The mistake	Why it is wrong	The fix
Writing $\lim_{x \to 2} \frac{1}{x-2} = \infty$ without specifying positive or negative infinity.	The one-sided limits differ ( $+\infty$ from the right, $-\infty$ from the left), so an unsigned statement is incomplete and masks the asymptote's directional behavior.	Always compute left and right limits separately when the denominator changes sign. Write $\lim_{x \to 2^{+}} \frac{1}{x-2} = +\infty$ and $\lim_{x \to 2^{-}} \frac{1}{x-2} = -\infty$ , or state that the two-sided limit does not exist.
Claiming $\lim_{x \to 0} \frac{1}{x^2} = -\infty$ because the denominator is zero.	Even though the denominator is zero, $x^2$ is always positive, so a positive numerator divided by a positive denominator approaching zero can never yield $-\infty$ .	Always track the sign of the denominator from both directions. Since $x^2 > 0$ for all $x \ne 0$ , the correct answer is $\lim_{x \to 0} \frac{1}{x^2} = +\infty$ .
Treating $\lim_{x \to 3} \frac{x-3}{x-3}$ as an infinite limit because the denominator approaches zero.	The numerator and denominator have a common factor that cancels for all $x \ne 3$ , so the simplified function is 1—a finite number with no infinite limit.	Simplify algebraically before applying limit rules: $\frac{x-3}{x-3} = 1$ for $x \ne 3$ , so $\lim_{x \to 3} \frac{x-3}{x-3} = 1$ . Infinite limits only occur when a zero in the denominator cannot be canceled by the numerator.

Tips and when to use something else

Check the sign of the numerator and denominator on both sides of the target value—if the denominator changes sign but the numerator does not, the left and right infinite limits will have opposite signs.
Look for common factors in the numerator and denominator before concluding the limit is infinite—factoring and canceling may eliminate the zero denominator entirely.
Infinite limits describe vertical asymptotes at a finite value $x = a$ ; if instead you see $x \to \infty$ or $x \to -\infty$ , use Limits at Infinity instead to describe horizontal asymptotes.
Build intuition by testing numerical values close to the target—substitute $x = a + 0.1$ , $x = a - 0.1$ , $x = a + 0.01$ to see whether the function output is growing positively, negatively, or staying bounded.

Related formulas

Definition of a LimitDefinition of a Limit describes what value a function approaches as its input approaches a specific point—essential for all of calculus.One-Sided LimitsOne-sided limits let you examine how a function behaves as you approach a point from just the left or right, determining whether a two-sided limit exists.Asymptotes from LimitsUse limits at infinity to determine where a function is heading as x grows without bound, then identify that destination as a horizontal asymptote equation.Limits at InfinityFind what a rational function approaches as x grows indefinitely by comparing the degrees and leading coefficients of the numerator and denominator.

Frequently asked questions

What's the difference between $\lim_{x \to 3} f(x) = \infty$ and $f(3) = \infty$ ?: The limit describes behavior as $x$ approaches 3, not what happens exactly at 3. Since infinity is not a real number, $f(3) = \infty$ is undefined—the function typically has no output at $x = 3$ . However, $\lim_{x \to 3} f(x)$ can still equal $\infty$ . For instance, $f(x) = \frac{1}{x-3}$ is undefined at $x = 3$ , yet the one-sided limits are infinite.
Can one side of a limit be infinite while the other is finite?: No; if either one-sided limit is infinite, the two-sided limit does not exist. However, you can have one one-sided limit be $+\infty$ and the other be $-\infty$ , as in $\lim_{x \to 0^{-}} \frac{1}{x} = -\infty$ and $\lim_{x \to 0^{+}} \frac{1}{x} = +\infty$ . In this case, the two-sided limit $\lim_{x \to 0} \frac{1}{x}$ is undefined.
How do I know if a limit is $+\infty$ or $-\infty$ just by looking at the expression?: Factor the numerator and denominator, then determine the sign of each piece as $x$ approaches the target value from each direction. The product of the numerator's sign and the denominator's sign tells you whether the result is $+\infty$ or $-\infty$ . Creating a sign chart or testing points slightly to the left and right of the target value helps determine signs accurately.
Are infinite limits the same as discontinuities?: Infinite limits indicate a vertical asymptote, which is one specific type of discontinuity, but not all discontinuities are infinite. A jump discontinuity (where left and right limits are finite but different) or a removable discontinuity (a hole in the graph) do not involve infinite limits. Functions with infinite limits are always discontinuous, but discontinuity does not always mean an infinite limit exists.

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Reviewed 2026-09-18