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Evaluating Limits



Direct Substitution — Try This First
When Direct Substitution Fails
Indeterminate Forms
Notation of Indeterminate Forms
Factoring and Canceling
Rationalizing — Conjugate Multiplication
Expanding and Simplifying
Combining Fractions
Multiplying by Strategic Forms of 1
Using Known Limits
One-Sided Evaluation
Sign Analysis Near the Point
Worked Examples
Summary: The Limit Evaluation Playbook
Evaluating Limits FAQ



When Substitution Fails


The simplest approach to any limit is direct substitution: plug in the value and compute. For polynomials, this always works. For rational functions away from zeros of the denominator, it works just as well. But substitution has limits of its own.

When plugging in produces 0/00/0, the expression is indeterminate—neither the numerator nor denominator alone determines the result. The limit might be any finite number, or infinite, or nonexistent. The form 0/00/0 signals that cancellation is hiding the true behavior, and algebraic work is required to reveal it.

This page covers the core techniques: factoring, rationalizing, and algebraic manipulation. Each method transforms an indeterminate expression into one where substitution succeeds.

Key Terms

Limit— the value being computed
Indeterminate Form— 00\frac{0}{0} and related forms that require algebraic resolution
Continuity— when present, direct substitution gives the limit
One-Sided Limit— needed when left and right behavior differ

See All Calculus Definitions →


Direct Substitution — Try This First


    Before attempting any technique, substitute aa into f(x)f(x) directly. If the result is a finite number with no division by zero, that number is the limit:

    Direct Substitution — Try This First
    lim⁡x→af(x)=f(a)\lim_{x \to a} f(x) = f(a)

    This works for:

  • Polynomials — always
  • Rational functions — when the denominator is nonzero at aa
  • Exponential, logarithmic, and trigonometric functions — at points in their domains
  • Sums, products, and compositions of continuous functions

    • Direct substitution exploits continuity. When ff is continuous at aa, the limit equals the function value by definition.
Function class Direct substitution gives the limit when…
Polynomial always — polynomials are continuous everywhere
Rational p(x) / q(x) q(a) ≠ 0
Exponential ax (a > 0) always — continuous everywhere
Logarithm ln x (and logb x) a > 0 — defined only on the positive reals
Trig: sin x, cos x always — continuous everywhere
Trig: tan x, sec x, csc x, cot x a is not a vertical asymptote of the function
Sums, products, and compositions of the above each piece is continuous at the relevant input

When Direct Substitution Fails


Substitution fails when it produces an undefined or indeterminate expression. The most common outcome is 0/00/0: both numerator and denominator evaluate to zero.

For example:

When Direct Substitution Fails
lim⁡x→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}

Substituting x=2x = 2 gives 00\dfrac{0}{0}. This does not mean the limit is zero, undefined, or nonexistent. It means the expression's behavior near x=2x = 2 is not yet determined—more work is needed.

The 0/00/0 form indicates that both numerator and denominator share a common factor of (x−2)(x - 2). Removing this factor reveals the limit.
(x^2 - 4)/(x - 2) is the line x + 2 with a hole at (2, 4): substitution gives 0/0 but the limit is 4xy−112345−11234567y = (x² − 4)/(x − 2)at x = 2: 0/0, no valueheights near x = 2 → 40/0 at the point, yet the limit is 4: the graph has only a hole
What 0/0 hides

The section’s (x² − 4)/(x − 2). For every x ≠ 2 it equals x + 2, so its graph is that line with a single point missing at (2, 4): substituting x = 2 gives 0/0 because the formula has no value there. The heights on both sides still close in on 4, and the limit is 4. 0/0 says only that more work is needed, which factoring and canceling then supplies.

Factoring in section 4 removes the common factor x − 2 and fills in the value 4.

Indeterminate Forms


Several forms signal that limit rules cannot be applied directly:

Indeterminate Forms
00∞∞0⋅∞∞−∞\frac{0}{0} \qquad \frac{\infty}{\infty} \qquad 0 \cdot \infty \qquad \infty - \infty

Indeterminate Forms
001∞∞00^0 \qquad 1^\infty \qquad \infty^0

Each form represents a competition between opposing tendencies. In 0/00/0, both numerator and denominator vanish—which vanishes faster determines the limit. In ∞−∞\infty - \infty, both terms grow without bound—their difference depends on relative growth rates.

Indeterminate forms require transformation. The goal is to rewrite the expression so that substitution or limit rules apply.

Notation of Indeterminate Forms

Notation

Notation of Indeterminate Forms

The seven form-labels, the lookalikes that are not on the list, and the one piece of working notation every evaluation chain depends on.
lim⁡\lim and DNE — two-sided limits; ∞\infty and its signs — limits at infinity.
00\frac{0}{0} · ∞∞\frac{\infty}{\infty}
Zero over zero; infinity over infinity
Diagnoses, not fractions. Each records what substitution reported and announces that the contest between numerator and denominator is still undecided. No arithmetic is being performed.
Cases0/00/0 at a finite point — a shared factor is hiding, removed in Factoring and Canceling below. ∞/∞\infty/\infty — a growth-rate contest, settled by dominant terms on limits at infinity or by L'Hôpital once derivative rules are available.
Also writtenQuoted — “the form 0/00/0” — in careful texts, stressing label over value.
Do not confuseAn answer. 0/00/0 does not mean the limit is 00, undefined, or nonexistent — it means undetermined, and any of those outcomes is still possible.
0⋅∞0 \cdot \infty · ∞−∞\infty - \infty · 000^0 · 1∞1^{\infty} · ∞0\infty^0
The product, difference, and power indeterminate forms
The same undecided-contest idea in three more shapes: a vanishing factor against an exploding one, two explosions subtracted, and exponentials pulling in opposite directions.
Cases0⋅∞0 \cdot \infty — rewrite as a quotient to join the entry above. ∞−∞\infty - \infty — combine into one fraction. The power trio — take logarithms, which turns each into 0⋅∞0 \cdot \infty.
Also written∞⋅0\infty \cdot 0, order immaterial.
Do not confuse1∞1^{\infty} with plain powers of one. The base is not exactly 11 — it approaches 11 while the exponent explodes, which is how (1+1n)n\left(1 + \frac{1}{n}\right)^n reaches ee on special limits.
10\frac{1}{0} · 0∞\frac{0}{\infty} · ∞+∞\infty + \infty
The determinate lookalikes
Not on the list. In each, one tendency wins outright: 1/01/0 forces unboundedness, 0/∞0/\infty forces 00, ∞+∞\infty + \infty forces ∞\infty. No contest, no further work — only sign analysis remains.
Cases1/01/0 resolves to an infinite limit once the sign is read off each side, as in Sign Analysis Near the Point below.
Do not confuseMembership in the indeterminate list. Students promote 1/01/0 to indeterminate because it fails substitution — but failing substitution and being undecided are different events; 1/01/0 has a verdict.
lim⁡⋯=lim⁡⋯=L\lim \cdots = \lim \cdots = L
The evaluation chain
Working notation: every intermediate line keeps its own lim⁡\lim. The algebra rewrites the expression inside the operator; lim⁡\lim drops only at the substitution step, when a number finally appears.
CasesThe cancellations along the chain are legal because x≠ax \neq a throughout — the punctured 0<∣x−a∣0 < |x - a| carried by every limit.
Do not confuseDropping lim⁡\lim after the first line. The naked middle lines then assert that x2−4x−2\frac{x^2 - 4}{x - 2} equals x+2x + 2 — false at x=2x = 2; only their limits agree. The commonest written error in submitted work.

Factoring and Canceling


When substitution yields 0/00/0, the numerator and denominator share a common factor. Factor both, cancel the shared factor, then substitute.

Factoring and Canceling
lim⁡x→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}

Factor the numerator:

Factoring and Canceling
=lim⁡x→2(x−2)(x+2)x−2= \lim_{x \to 2} \frac{(x - 2)(x + 2)}{x - 2}

Cancel (x−2)(x - 2):

Factoring and Canceling
=lim⁡x→2(x+2)=4= \lim_{x \to 2} (x + 2) = 4

The cancellation is valid because the limit considers xx near 22, not at 22. For x≠2x \neq 2, the factor (x−2)(x - 2) is nonzero and cancels legitimately.

Rationalizing — Conjugate Multiplication


When radicals appear and substitution fails, multiply by the conjugate to eliminate the radical.

Rationalizing — Conjugate Multiplication
lim⁡x→0x+1−1x\lim_{x \to 0} \frac{\sqrt{x + 1} - 1}{x}

Substituting gives 0/00/0. Multiply numerator and denominator by x+1+1\sqrt{x + 1} + 1:

Rationalizing — Conjugate Multiplication
=lim⁡x→0(x+1−1)(x+1+1)x(x+1+1)= \lim_{x \to 0} \frac{(\sqrt{x + 1} - 1)(\sqrt{x + 1} + 1)}{x(\sqrt{x + 1} + 1)}

The numerator becomes a difference of squares:

Rationalizing — Conjugate Multiplication
=lim⁡x→0(x+1)−1x(x+1+1)=lim⁡x→0xx(x+1+1)= \lim_{x \to 0} \frac{(x + 1) - 1}{x(\sqrt{x + 1} + 1)} = \lim_{x \to 0} \frac{x}{x(\sqrt{x + 1} + 1)}

Cancel xx:

Rationalizing — Conjugate Multiplication
=lim⁡x→01x+1+1=12= \lim_{x \to 0} \frac{1}{\sqrt{x + 1} + 1} = \frac{1}{2}

Expanding and Simplifying


Sometimes expanding a product or simplifying a complex fraction reveals the cancellation needed.

Expanding and Simplifying
lim⁡x→1(x+1)2−4x−1\lim_{x \to 1} \frac{(x + 1)^2 - 4}{x - 1}

Expand the numerator:

Expanding and Simplifying
=lim⁡x→1x2+2x+1−4x−1=lim⁡x→1x2+2x−3x−1= \lim_{x \to 1} \frac{x^2 + 2x + 1 - 4}{x - 1} = \lim_{x \to 1} \frac{x^2 + 2x - 3}{x - 1}

Factor:

Expanding and Simplifying
=lim⁡x→1(x−1)(x+3)x−1=lim⁡x→1(x+3)=4= \lim_{x \to 1} \frac{(x - 1)(x + 3)}{x - 1} = \lim_{x \to 1} (x + 3) = 4

The initial form obscured the factor of (x−1)(x - 1); expansion made it visible.

Combining Fractions


When the expression involves a difference of fractions, combine them over a common denominator.

Combining Fractions
lim⁡x→1(1x−1−2x2−1)\lim_{x \to 1} \left( \frac{1}{x - 1} - \frac{2}{x^2 - 1} \right)

Note that x2−1=(x−1)(x+1)x^2 - 1 = (x - 1)(x + 1). Rewrite with common denominator:

Combining Fractions
=lim⁡x→1(x+1(x−1)(x+1)−2(x−1)(x+1))= \lim_{x \to 1} \left( \frac{x + 1}{(x - 1)(x + 1)} - \frac{2}{(x - 1)(x + 1)} \right)

Combining Fractions
=lim⁡x→1x+1−2(x−1)(x+1)=lim⁡x→1x−1(x−1)(x+1)= \lim_{x \to 1} \frac{x + 1 - 2}{(x - 1)(x + 1)} = \lim_{x \to 1} \frac{x - 1}{(x - 1)(x + 1)}

Cancel (x−1)(x - 1):

Combining Fractions
=lim⁡x→11x+1=12= \lim_{x \to 1} \frac{1}{x + 1} = \frac{1}{2}

Multiplying by Strategic Forms of 1


For limits at infinity, divide numerator and denominator by the highest power of xx in the denominator.

Multiplying by Strategic Forms of 1
lim⁡x→∞3x2+5x−12x2−7\lim_{x \to \infty} \frac{3x^2 + 5x - 1}{2x^2 - 7}

Divide every term by x2x^2:

Multiplying by Strategic Forms of 1
=lim⁡x→∞3+5x−1x22−7x2= \lim_{x \to \infty} \frac{3 + \frac{5}{x} - \frac{1}{x^2}}{2 - \frac{7}{x^2}}

As x→∞x \to \infty, terms with xx in the denominator vanish:

Multiplying by Strategic Forms of 1
=3+0−02−0=32= \frac{3 + 0 - 0}{2 - 0} = \frac{3}{2}

This technique isolates the dominant terms that control behavior at infinity.

Using Known Limits


Recognize when parts of an expression match special limits and rewrite accordingly.

Using Known Limits
lim⁡x→0sin⁡3xx\lim_{x \to 0} \frac{\sin 3x}{x}

Rewrite to match the standard form sin⁡uu\dfrac{\sin u}{u}:

Using Known Limits
=lim⁡x→0sin⁡3xx⋅33=3⋅lim⁡x→0sin⁡3x3x= \lim_{x \to 0} \frac{\sin 3x}{x} \cdot \frac{3}{3} = 3 \cdot \lim_{x \to 0} \frac{\sin 3x}{3x}

Let u=3xu = 3x. As x→0x \to 0, u→0u \to 0:

Using Known Limits
=3⋅lim⁡u→0sin⁡uu=3⋅1=3= 3 \cdot \lim_{u \to 0} \frac{\sin u}{u} = 3 \cdot 1 = 3

One-Sided Evaluation


When one-sided limits differ, evaluate each separately.

One-Sided Evaluation
lim⁡x→0∣x∣x\lim_{x \to 0} \frac{|x|}{x}

For x>0x > 0: ∣x∣=x|x| = x, so ∣x∣x=1\dfrac{|x|}{x} = 1

For x<0x < 0: ∣x∣=−x|x| = -x, so ∣x∣x=−1\dfrac{|x|}{x} = -1

One-Sided Evaluation
lim⁡x→0+∣x∣x=1lim⁡x→0−∣x∣x=−1\lim_{x \to 0^+} \frac{|x|}{x} = 1 \qquad \lim_{x \to 0^-} \frac{|x|}{x} = -1

The one-sided limits differ, so the two-sided limit does not exist.
-4 -3 -2 -1 1 2 3 4 -4 -3 -2 -1 1 2 3 4 x y f(x) = x for x<0; x+1 for x≥0
Step at x = 0: left limit 0, right limit 1

Evaluated from the left the probe heads to 0, from the right to 1; the two one-sided limits are perfectly well defined and simply differ, so the two-sided limit does not exist. Any expression that changes formula at a point, like |x|/x, has to be evaluated this way. Read each side separately on the limit explorer.

When the two sides agree the one-sided evaluation was unnecessary; when they differ it was the whole answer.

Sign Analysis Near the Point


When a limit involves potential division by zero with a nonzero numerator, determine the sign to identify +∞+\infty or −∞-\infty.

Sign Analysis Near the Point
lim⁡x→3+x+1x−3\lim_{x \to 3^+} \frac{x + 1}{x - 3}

At x=3x = 3: numerator =4>0= 4 > 0, denominator →0\to 0.

For xx slightly greater than 33: x−3>0x - 3 > 0 (small positive).

Positive divided by small positive gives large positive:

Sign Analysis Near the Point
lim⁡x→3+x+1x−3=+∞\lim_{x \to 3^+} \frac{x + 1}{x - 3} = +\infty

From the left, x−3<0x - 3 < 0, so:

Sign Analysis Near the Point
lim⁡x→3−x+1x−3=−∞\lim_{x \to 3^-} \frac{x + 1}{x - 3} = -\infty
(x + 1)/(x - 3): just right of 3 the quotient is large positive, just left large negativexy−2−112345678−12−8−44812x = 3y = (x + 1)/(x − 3)4 / (small +) → +∞4 / (small −) → −∞The sign of the small denominator decides +∞ or −∞
Which infinity, from which side

The section’s (x + 1)/(x − 3) near x = 3. The numerator stays near 4 while the denominator shrinks to 0. Just right of 3 the denominator is a small positive number, so the quotient is a large positive number and the graph shoots up; just left of 3 it is a small negative number and the graph drops. So the right-hand limit is +∞ and the left-hand limit is −∞.

Because the two sides disagree, the two-sided limit at 3 does not exist, not even as ∞.
Numerator sign at a Denominator sign as x → a Resulting limit
positive (+) small positive (0⁺) +∞
positive (+) small negative (0⁻) −∞
negative (−) small positive (0⁺) −∞
negative (−) small negative (0⁻) +∞

Worked Examples


Example 1: Factoring


Worked Examples
lim⁡x→3x2−9x−3=lim⁡x→3(x−3)(x+3)x−3=lim⁡x→3(x+3)=6\lim_{x \to 3} \frac{x^2 - 9}{x - 3} = \lim_{x \to 3} \frac{(x-3)(x+3)}{x-3} = \lim_{x \to 3}(x + 3) = 6

Example 2: Rationalizing


Worked Examples
lim⁡x→4x−2x−4=lim⁡x→4(x−2)(x+2)(x−4)(x+2)=lim⁡x→4x−4(x−4)(x+2)=14\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4} = \lim_{x \to 4} \frac{(\sqrt{x} - 2)(\sqrt{x} + 2)}{(x - 4)(\sqrt{x} + 2)} = \lim_{x \to 4} \frac{x - 4}{(x - 4)(\sqrt{x} + 2)} = \frac{1}{4}

Example 3: Using Special Limits


Cosine Quadratic Limit at Zero
lim⁡x→01−cos⁡xx2=12\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}
Learn more about this formula: Cosine Quadratic Limit at Zero →


This follows from the special limit lim⁡x→01−cos⁡xx2=12\lim_{x \to 0} \dfrac{1 - \cos x}{x^2} = \dfrac{1}{2}.

Example 4: Combining Fractions


Worked Examples
lim⁡x→2(1x−2−4x2−4)=lim⁡x→2x+2−4(x−2)(x+2)=lim⁡x→2x−2(x−2)(x+2)=14\lim_{x \to 2} \left( \frac{1}{x - 2} - \frac{4}{x^2 - 4} \right) = \lim_{x \to 2} \frac{x + 2 - 4}{(x-2)(x+2)} = \lim_{x \to 2} \frac{x - 2}{(x-2)(x+2)} = \frac{1}{4}

Summary: The Limit Evaluation Playbook


Every limit evaluation problem on this page follows the same arc: try direct substitution first, then classify the indeterminate form, then pick the technique that resolves it. The table below collects the nine moves in priority order — start at the top and move down only as needed. The right side of each row notes where on this page (or which sibling page) the move is detailed.
Step Trigger Technique Section
1 any limit — always start here direct substitution; if the result is defined and finite, that is the limit obj1
2 0/0 with a polynomial numerator and denominator factor both, cancel the shared (x − a) factor, substitute again obj4
3 0/0 with radicals (square roots, etc.) multiply numerator and denominator by the conjugate, simplify the difference of squares, cancel obj5
4 0/0 with shared factor hidden inside parens or powers expand or simplify the algebra until the (x − a) factor emerges, then cancel obj6
5 0/0 with a difference of fractions combine over a common denominator; the simplified numerator usually reveals the shared factor obj7
6 ∞/∞ for a rational function as x → ±∞ divide every term by the highest power of x in the denominator — see limits and infinity obj8
7 expression resembles a known trig/exp form rewrite to match a special limit (sin u / u, (eu−1)/u, etc.) and apply obj9
8 absolute values, piecewise functions, or potential left/right asymmetry evaluate the one-sided limits separately; agreement determines whether the two-sided limit exists obj10
9 nonzero / 0 (not indeterminate — limit is infinite) sign-analyze numerator and denominator near a from each side to identify +∞ or −∞ obj11

Evaluating Limits FAQ

What does 0/0 mean when evaluating a limit?

+
It is a diagnosis, not an answer. Substitution produced 0/0, which records only that the contest between numerator and denominator is still undecided — no arithmetic has been performed. The limit may yet turn out to be any number, or infinite, or nonexistent. Usually a shared factor is hiding in the expression, and factoring then canceling it resolves the form.Read more →

Is 1/0 an indeterminate form?

+
No. Indeterminate means undecided, and 1/0 has a verdict: a nonzero numerator over a vanishing denominator forces the magnitude to grow without bound. Only the sign is left to settle, by checking each side separately. Students often promote 1/0 to the indeterminate list because it fails direct substitution, but failing substitution and being undecided are different events.Read more →

Why is 1 to the power of infinity indeterminate?

+
Because the base is never exactly 1 — it only approaches 1 while the exponent explodes, and the two tendencies pull against each other. A base slightly above 1 raised to a huge power grows without bound; slightly below 1, it collapses toward zero. That balance is why (1 + 1/n) to the n approaches e rather than 1. See special limits.Read more →

Do you need to write lim on every line?

+
Yes, until the substitution step. The algebra rewrites the expression inside the operator, so each intermediate line keeps its own lim; the symbol drops only once a number finally appears. Omitting it makes the middle lines claim that (x² − 4)/(x − 2) equals x + 2, which is false at x = 2 — only their limits agree.Read more →