Limits — The Idea Behind lim x→a f(x)
The limit is the first genuinely new idea in calculus, and the one most students skip past. They learn "substitute the value", and it works — until the substitution gives 0/0 and the whole method collapses. This article explains what a limit actually asks, what a one-sided limit is, why a limit can exist where the function has no value at all, and how to handle each indeterminate form with the arithmetic written out in full.
What a limit actually asks
The statement
does not mean "f(a) = L". It means something narrower: as x gets closer and closer to a — from either side, but never equal to a — the value f(x) settles as close as we like to the number L.
The phrase never equal to a is the whole point. A limit deliberately ignores what happens at x = a and looks only at the neighbourhood around it. That single restriction is what lets calculus make sense of expressions like 0/0.
Seeing it in a table of values
Take f(x) = (x² − 4) / (x − 2). At x = 2 the formula gives (4 − 4)/(2 − 2) = 0/0, which is meaningless — so f(2) simply does not exist. Now watch what the nearby values do:
| x | f(x) = (x² − 4)/(x − 2) |
|---|---|
| 1.9 | (3.61 − 4)/(−0.1) = (−0.39)/(−0.1) = 3.9 |
| 1.99 | (3.9601 − 4)/(−0.01) = 3.99 |
| 2 | undefined (0/0) |
| 2.01 | (4.0401 − 4)/(0.01) = 4.01 |
| 2.1 | (4.41 − 4)/(0.1) = 0.41/0.1 = 4.1 |
From both sides the values crowd in on 4, so the limit is 4 even though the function has a hole at x = 2. A limit can exist where the function does not — that is the reason limits were invented.
One-sided limits
Sometimes the two sides disagree. We write them separately:
Right-hand limit (RHL): limx→a⁺ f(x) — approach a from values larger than a
The limit limx→a f(x) exists only if LHL = RHL, and both are finite.
Test it on f(x) = |x − 3| / (x − 3):
- For x > 3, the quantity x − 3 is positive, so |x − 3| = x − 3 and f(x) = 1. Therefore RHL = 1.
- For x < 3, the quantity x − 3 is negative, so |x − 3| = −(x − 3) and f(x) = −1. Therefore LHL = −1.
Since −1 ≠ 1, the limit at x = 3 does not exist. The graph jumps. Any function with a modulus, a greatest-integer part or a piecewise definition must be checked from both sides.
The indeterminate forms
An indeterminate form is a substitution result that carries no information — the answer could be anything until you do more work. These are the seven you will meet:
| Form | Usual method |
|---|---|
| 0 / 0 | Factorise and cancel; or rationalise; or use a standard limit |
| ∞ / ∞ | Divide numerator and denominator by the highest power of x |
| ∞ − ∞ | Combine into a single fraction, or rationalise, then re-check the form |
| 0 × ∞ | Rewrite as a fraction so it becomes 0/0 or ∞/∞ |
| 1∞ | Use the standard result limx→∞ (1 + 1/x)x = e |
| 00 | Take logarithms, evaluate, then exponentiate back |
| ∞0 | Take logarithms, evaluate, then exponentiate back |
Note what is not on the list. A result like 5/0 is not indeterminate — it says the function is blowing up. And 0/5 is simply 0. Only these seven need extra work.
Worked example 1 — the 0/0 form, by factorising
Evaluate limx→3 (x² − 9) / (x² − x − 6).
Step 1 — check the form. Numerator at x = 3: 9 − 9 = 0. Denominator at x = 3: 9 − 3 − 6 = 0. So it is 0/0 — indeterminate, work needed.
Step 2 — factorise both.
x² − 9 = (x − 3)(x + 3)
x² − x − 6 = (x − 3)(x + 2)
Step 3 — cancel (x − 3). This is legal because x ≠ 3 inside the limit — x only approaches 3. Write that line down; examiners look for it.
= limx→3 (x + 3)/(x + 2) = (3 + 3)/(3 + 2) = 6/5 = 1.2
Check at x = 3.001: 0.006001 / 0.005001 = 1.19996 ✔
Worked example 2 — the 0/0 form, by rationalising
Evaluate limx→0 (√(x + 9) − 3) / x.
Step 1 — check the form. At x = 0: (√9 − 3)/0 = (3 − 3)/0 = 0/0.
Step 2 — multiply above and below by the conjugate (√(x + 9) + 3):
numerator becomes (x + 9) − 9 = x, so the expression is
x / [ x (√(x + 9) + 3) ]
Step 3 — cancel x (again valid since x ≠ 0):
= limx→0 1 / (√(x + 9) + 3) = 1/(3 + 3) = 1/6 ≈ 0.1667
Numerical check at x = 0.01: √9.01 = 3.0016662, so (3.0016662 − 3)/0.01 = 0.0016662/0.01 = 0.16662 ✔
Worked example 3 — the ∞/∞ form
Evaluate limx→∞ (3x² + 5x − 2) / (2x² − x + 7).
Step 1 — check the form. Both top and bottom grow without bound: ∞/∞.
Step 2 — divide every term by x² (the highest power present):
= limx→∞ (3 + 5/x − 2/x²) / (2 − 1/x + 7/x²)
Step 3 — as x → ∞ every term with x in the denominator goes to 0:
= (3 + 0 − 0)/(2 − 0 + 0) = 3/2 = 1.5
Numerical check at x = 1000: numerator = 3 000 000 + 5000 − 2 = 3 004 998; denominator = 2 000 000 − 1000 + 7 = 1 999 007; ratio = 1.5033 — already close, and it keeps closing in ✔
Worked example 4 — the standard limit sin x / x
Evaluate limx→0 sin 5x / (3x).
Substituting gives 0/0. Use the standard result limθ→0 (sin θ)/θ = 1, with x in radians. Force the matching angle by multiplying and dividing by 5x:
sin 5x / (3x) = (sin 5x / 5x) × (5x / 3x) = (sin 5x / 5x) × (5/3)
As x → 0, 5x → 0 too, so the first bracket → 1, giving 1 × 5/3 = 5/3 ≈ 1.6667
Numerical check at x = 0.001 rad: sin(0.005) = 0.00499998; 3x = 0.003; ratio = 0.00499998/0.003 = 1.66666 ✔
Standard limits worth memorising
| Limit | Value |
|---|---|
| limx→a (xⁿ − aⁿ)/(x − a) | n an−1 |
| limx→0 (sin x)/x (radians) | 1 |
| limx→0 (tan x)/x (radians) | 1 |
| limx→0 (1 − cos x)/x² | 1/2 |
| limx→0 (ex − 1)/x | 1 |
| limx→0 loge(1 + x)/x | 1 |
| limx→∞ (1 + 1/x)x | e ≈ 2.71828 |
The first entry is worth staring at: with f(x) = xⁿ it is exactly the definition of the derivative at x = a.
Common mistakes that cost marks
- Writing 0/0 = 1 or 0/0 = 0. It is neither. It is a signal that the expression must be simplified first — the answer could be any number, as examples 1 and 2 show (1.2 and 0.1667 both came from 0/0).
- Cancelling without stating the reason. You may cancel (x − 3) only because x ≠ 3 inside a limit. Write it; it carries marks in step-marking schemes.
- Using degrees in trigonometric limits. (sin x)/x → 1 is true only in radians. In degrees, sin(0.1°) = 0.0017453, so the ratio is about 0.01745 — that is π/180.
- Assuming the limit equals f(a). That shortcut (called continuity) works for polynomials, but not at a hole, a jump or an asymptote — check the form first.
- Ignoring one side. For modulus, greatest-integer and piecewise functions, compute LHL and RHL separately and compare them.
- Treating ∞ as a number. You cannot write ∞ − ∞ = 0.
Where limits appear in exams
| Exam / subject | Typical use |
|---|---|
| CBSE/ICSE Class 11 | Limits and Derivatives — algebraic and trigonometric limits, first principles |
| CBSE/ICSE Class 12 | Continuity and differentiability — LHL, RHL and f(a) compared at a point |
| JEE Main & Advanced | Indeterminate forms, expansions, limits of sequences and series |
| Physics and physical chemistry | Instantaneous velocity, and instantaneous rate as the limiting slope of a concentration–time curve |
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