MATHEMATICS · LESSON 02 OF 10

Limits

A limit describes the value a function approaches near a point, not the value at the point.

What this lesson explains

Every idea in calculus is built on limits. The derivative (instantaneous rate of change) is a limit of average rates, and continuity is defined with a limit. Engineers use limits whenever a quantity cannot be evaluated directly at a point but its behaviour close to that point is known: the speed at one instant, the steady value a process settles to, or a formula that breaks down at a single input.

Before you begin

Factoring

Difference of squares: a² − b² = (a − b)(a + b), so x² − 9 = (x − 3)(x + 3). A quadratic x² + bx + c factors as (x + p)(x + q) when p + q = b and pq = c; for example x² − 5x + 6 = (x − 2)(x − 3).

Cancelling in a fraction

A common factor may be cancelled only when it is not zero. The statement (x − 3)(x + 3) / x − 3 = x + 3 is true for every x except x = 3. A limit never uses the input x = a itself, so inside a limit the cancellation is valid.

Conjugates

Multiplying √A − B by its conjugate √A + B gives A − B², which removes the square root. Multiplying the numerator and the denominator by the same non-zero expression does not change the value of a fraction.

The idea, made visible

Think of walking along the graph of a function towards the input x = a. The limit asks: what height are you heading towards? It does not ask what happens exactly at x = a. The function may have a different value there, or no value at all, and the limit is unaffected.

For example, f(x) = x² − 9 / x − 3 cannot be evaluated at x = 3, because the denominator is zero. For every other input, the numerator factors as (x − 3)(x + 3), so f(x) = x + 3 whenever x ≠ 3. Near 3 the outputs are close to 6, so the limit is 6, even though f(3) does not exist. The graph is a straight line with a single hole.

The graph of (x² − 9)/(x − 3) is the line y = x + 3 with one point removed. The heights approach 6 from both sides, so the limit is 6 even though there is no value at x = 3.

You approach a point from two sides. Coming from inputs smaller than a gives the left-hand limit; coming from larger inputs gives the right-hand limit. The (two-sided) limit exists only when both one-sided limits exist and agree.

When direct substitution produces the expression 0 / 0, the calculation is not finished: 0 / 0 is not a number. Often it signals that the numerator and denominator share a factor that vanishes at the point; in other cases, such as sin x / x at 0, a known limit is needed instead. Either way, rewrite the expression, using steps that are valid for inputs near a (but not at a), until the limit can be read off.

Values of f(x) = (x² − 9)/(x − 3) near x = 3
x	side	f(x)
2.9	left	5.9
2.99	left	5.99
2.999	left	5.999
3	the point itself	undefined (0/0)
3.001	right	6.001
3.01	right	6.01
3.1	right	6.1

Both columns of nearby values head towards 6. The middle row has no value, yet the limit is 6: a limit is decided by the surrounding rows. A table suggests a limit; algebra confirms it.

Key terms

Limit: We write lim_x→a f(x) = L when the values f(x) can be made as close to L as we like by taking x sufficiently close to a, with x ≠ a. The value f(a), if it exists, plays no part.
One-sided limits: lim_x→a⁻ f(x) uses only inputs x < a; lim_x→a⁺ f(x) uses only x > a. The two-sided limit equals L exactly when both one-sided limits equal L.
Indeterminate form: An expression such as 0 / 0 obtained by direct substitution. It tells you that more work is needed; different functions giving 0 / 0 can have completely different limits.
Limit at infinity: lim_x→∞ f(x) = L means f(x) gets arbitrarily close to L as x grows without bound. The line y = L is then a horizontal asymptote of the graph.
Infinite limit: lim_x→a f(x) = ∞ means f(x) grows without bound near a. The limit does not exist as a number; writing ∞ describes how it fails. The line x = a is a vertical asymptote.

The formulas and what they mean

Symbols, meanings and units
Symbol	Meaning	Unit
x → a	x approaches a, taking values near a but never equal to a	same units as x
x → a⁻, x → a⁺	approach from the left (x < a) or from the right (x > a)	—
L	the limiting value of f(x)	same units as f(x)
ε, δ	epsilon: allowed error in the output; delta: allowed distance of the input from a	output units, input units

Existence test

lim_x→a f(x) = L ⇔ lim_x→a⁻ f(x) = L and lim_x→a⁺ f(x) = L

Conditions and limits: Always. If the one-sided limits differ, or either fails to exist, the two-sided limit does not exist.

Limit laws

lim (f ± g) = lim f ± lim g; lim (f·g) = (lim f)(lim g); lim f / g = lim f / lim g

Conditions and limits: Valid when lim f and lim g exist (as finite numbers), and for the quotient only when lim g ≠ 0.

Direct substitution

lim_x→a p(x) = p(a)

Conditions and limits: For polynomials at every a; for rational functions p / q wherever q(a) ≠ 0. Also for √x, sin x, cos x, e^x at points of their domain.

Fundamental trigonometric limit

lim_x→0 sin x / x = 1

Conditions and limits: x must be measured in radians. In degrees the limit is π/180, not 1.

Squeeze theorem

If g(x) ≤ f(x) ≤ h(x) near a and lim g = lim h = L, then lim f = L

Conditions and limits: The inequalities must hold for all x near a (except possibly at a).

Rational functions at infinity

lim_x→∞ a_nxⁿ + … / b_mx^m + … = 0 if n < m; a_n / b_m if n = m; ±∞ if n > m

Conditions and limits: Divide numerator and denominator by the highest power of x in the denominator. For n > m the sign depends on the leading coefficients and on whether x → +∞ or −∞.

Why it works: Why cancelling a factor that is zero at the point is allowed

Cancelling (x − 3) looks like dividing by zero at x = 3. The definition of a limit is what makes it legitimate.

Factor: x² − 9 / x − 3 = (x − 3)(x + 3) / x − 3.
Factoring exposes the factor that makes both numerator and denominator zero.
The limit only uses inputs with x ≠ 3.
By definition, lim_x→3 looks at x near 3 and excludes x = 3 itself.
For x ≠ 3, x − 3 ≠ 0, so the factor cancels: the expression equals x + 3.
Dividing by a non-zero number is always allowed.
The two functions agree at every input the limit uses, so they have the same limit: lim_x→3 (x + 3) = 6.
x + 3 is a polynomial, so direct substitution is valid for it.

The original expression is still undefined at x = 3. We have found its limit, not its value.

A first worked example

How to evaluate a limit lim_x→a f(x)

Try direct substitution first. If you get a number (and no division by zero), that number is the limit for polynomials, rational functions with non-zero denominator, roots, exponentials and trigonometric functions in their domains.
If you get 0 / 0: factor and cancel, or multiply by a conjugate when there is a square root, or use a known limit such as sin x / x → 1.
If you get k / 0 with k ≠ 0: the function grows without bound. Check the sign on each side to decide between +∞, −∞ or “does not exist”.
For piecewise functions or absolute values, compute the left- and right-hand limits separately and compare them.
For x → ±∞ with a rational function, divide by the highest power of x in the denominator.

Factoring a 0/0 form

Problem. Find lim_x→2 x² − 4 / x − 2.

Substitute x = 2: numerator 4 − 4 = 0, denominator 2 − 2 = 0. The form is 0 / 0.
Substitution first tells us whether more work is needed.
Factor the numerator: x² − 4 = (x − 2)(x + 2).
Zero in both parts means (x − 2) is a common factor.
Cancel (x − 2), valid because x ≠ 2 inside the limit: the expression becomes x + 2.
The limit never uses x = 2 itself.
Substitute into the polynomial: 2 + 2 = 4.
Direct substitution is valid for polynomials.

Result: lim_x→2 x² − 4 / x − 2 = 4.

What it means: The graph is the line y = x + 2 with a hole at (2, 4).

A different case

A square root: multiply by the conjugate

Problem. Find lim_x→0 √x + 4 − 2 / x.

Substitute x = 0: numerator √4 − 2 = 0, denominator 0. The form is 0 / 0.
Factoring does not help directly because of the root.
Multiply numerator and denominator by √x + 4 + 2.
(√A − 2)(√A + 2) = A − 4 removes the root.
Numerator: (x + 4) − 4 = x. The fraction becomes x / x(√x + 4 + 2).
The hidden common factor x now appears.
Cancel x (x ≠ 0): 1 / √x + 4 + 2. Substitute x = 0: 1 / 2 + 2 = 1 / 4.
The new denominator is 4, not zero, so substitution is valid.

Result: The limit is 1/4 = 0.25.

What it means: This limit is the slope of y = √x at x = 4, a preview of the derivative.

More worked cases

Each case below uses a different skill. Every step and result is shown.

Value versus limit

Problem. Let f(x) = x² − 4 / x − 2 for x ≠ 2 and f(2) = 7. Find lim_x→2 f(x) and f(2).

For x ≠ 2, f(x) = x + 2 (as in the first example).
Only inputs near 2, not equal to 2, matter for the limit.
lim_x→2 f(x) = 2 + 2 = 4.
Substitute into x + 2.
The function value is given separately: f(2) = 7.
The value at the point is a separate piece of information.

Result: lim_x→2 f(x) = 4, but f(2) = 7.

What it means: The limit and the value are different, so this function is not continuous at x = 2 (see the next topic).

One-sided limits that disagree

Problem. Let f(x) = ∣x − 5∣ / x − 5. Find the one-sided limits at x = 5 and decide whether lim_x→5 f(x) exists.

For x > 5, x − 5 is positive, so |x − 5| = x − 5 and f(x) = 1.
The absolute value of a positive number is the number itself.
For x < 5, x − 5 is negative, so |x − 5| = −(x − 5), which is positive. Then f(x) = −(x − 5) / x − 5 = −1.
The absolute value of a negative number is its opposite, which is positive.
lim_x→5⁻ f(x) = −1 and lim_x→5⁺ f(x) = 1.
Each side is constant.

Result: The one-sided limits are −1 and 1. They differ, so lim_x→5 f(x) does not exist.

What it means: The graph jumps from height −1 to height 1 at x = 5.

A trigonometric limit (radians)

Problem. Find lim_x→0 sin 5x / x.

Rewrite: sin 5x / x = 5 · sin 5x / 5x.
Multiply and divide by 5 so the argument of sine matches the denominator.
Let u = 5x. As x → 0, u → 0, so sin u / u → 1.
The fundamental limit, with angles in radians.

Result: lim_x→0 sin 5x / x = 5 · 1 = 5.

What it means: Near zero, sin 5x ≈ 5x, so the ratio is close to 5.

A limit at infinity

Problem. Find lim_x→∞ 6x² + 1 / 2x² − 3.

Divide numerator and denominator by x²: 6 + 1/x² / 2 − 3/x².
x² is the highest power in the denominator.
As x → ∞, 1/x² → 0 and 3/x² → 0.
A constant divided by an ever larger number approaches 0.

Result: The limit is 6/2 = 3, so y = 3 is a horizontal asymptote.

What it means: For equal degrees, the limit is the ratio of the leading coefficients.

The squeeze theorem

Problem. Find lim_x→0 x² cos(1/x).

For every x ≠ 0, −1 ≤ cos(1/x) ≤ 1.
Cosine is always between −1 and 1, even though cos(1/x) oscillates wildly near 0 and has no limit itself.
Multiply by x² ≥ 0: −x² ≤ x² cos(1/x) ≤ x².
Multiplying by a non-negative number keeps the inequalities in the same direction.
Both outer functions have limit 0 as x → 0.
Direct substitution in ±x².

Result: By the squeeze theorem, lim_x→0 x² cos(1/x) = 0.

What it means: The factor x² shrinks the oscillation to nothing.

Common misunderstandings

Misunderstanding: “Substitution gives 0/0, so the limit is 0 (or 1, or does not exist).”
Correct idea: 0/0 is not an answer. lim_x→0 x/x = 1, lim_x→0 x²/x = 0 and lim_x→0 5x/x = 5 all give 0/0 on substitution but have different limits.
Misunderstanding: “The limit is the same as the value f(a).”
Correct idea: Only for continuous functions. A function can have a limit where it is undefined (the hole above) or a value different from its limit.
Misunderstanding: “If the left-hand limit exists, the limit exists.”
Correct idea: You must check both sides. One side alone can prove that a limit does not exist (if it disagrees with the other), never that it exists.
Misunderstanding: Treating ∞ as a number, e.g. “∞ − ∞ = 0”.
Correct idea: Forms such as ∞ − ∞ and ∞/∞ are indeterminate. Rewrite the expression (for example divide by the highest power) before concluding.
Misunderstanding: Using degrees in lim_x→0 sin x / x.
Correct idea: The value 1 requires radians. Calculators set to degrees give about 0.01745.

Going further (optional): The precise (ε–δ) meaning of a limit

lim_x→a f(x) = L means: for every tolerance ε > 0 on the output there is a distance δ > 0 such that 0 < |x − a| < δ guarantees |f(x) − L| < ε. In words, however strict the output tolerance, we can meet it by keeping the input close enough to a.

Example: for f(x) = 2x + 1 at a = 3, L = 7. Then |f(x) − 7| = |2x − 6| = 2|x − 3|. To make this less than ε we need |x − 3| < ε/2, so δ = ε/2 works. For ε = 0.1, δ = 0.05. The factor 2 is the slope: the function doubles input errors, so the input tolerance must be half the output tolerance.

Keep in mind

lim_x→a f(x) = L ⇔ lim_x→a⁻ f(x) = L and lim_x→a⁺ f(x) = LExistence test
lim (f ± g) = lim f ± lim g; lim (f·g) = (lim f)(lim g); lim f / g = lim f / lim gLimit laws
lim_x→a p(x) = p(a)Direct substitution
lim_x→0 sin x / x = 1Fundamental trigonometric limit
If g(x) ≤ f(x) ≤ h(x) near a and lim g = lim h = L, then lim f = LSqueeze theorem
lim_x→∞ a_nxⁿ + … / b_mx^m + … = 0 if n < m; a_n / b_m if n = m; ±∞ if n > mRational functions at infinity

Scope of this lesson

The ε–δ definition is explained with a linear example; general ε–δ proofs belong to real analysis.
L’Hôpital’s rule is not used here. It needs derivatives and has its own conditions; it appears as an optional section of Differentiation rules. Check whether your lecturer includes it in MA 101.

Next: Continuity. A function is continuous at a point when its limit there exists and equals its value: no holes, jumps or breaks.

Original study text. Sources and credits.

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