What is a limit? (the intuitive idea)
Think of a point moving along a graph. We move x closer and closer to a number a. We watch the height f(x).
If f(x) gets closer and closer to one number L, we write lim(x→a) f(x) = L. Read it as “the limit of f(x) as x tends to a is L”.
Important: the limit does not care about the value at a. The function may even have a hole there. It only cares about values near a.
Left-hand and right-hand limits
Coming from smaller numbers (like 1.9, 1.99) gives the left-hand limit, written x → a⁻. Coming from bigger numbers (2.1, 2.01) gives the right-hand limit, x → a⁺.
The limit exists only when both are equal. If the graph jumps, the two sides differ, so there is no limit.
Rules (algebra) of limits
If lim f(x) and lim g(x) both exist as x → a, then:
- limit of a sum = sum of the limits
- limit of a difference = difference of the limits
- limit of a product = product of the limits
- limit of a quotient = quotient of the limits (the bottom limit must not be 0)
- limit of k·f(x) = k × limit of f(x)
These rules let us break a hard limit into small easy pieces.
Limits of polynomial functions
A polynomial has no holes and no jumps. So its limit is just its value.
Example: lim(x→2) (3x² − x + 1) = 3(4) − 2 + 1 = 11.
Limits of rational functions
A rational function is one polynomial divided by another, p(x)/q(x).
- If q(a) ≠ 0, just put x = a.
- If you get 0/0, this is not an answer. It means (x − a) is a factor of both top and bottom. Factorise, cancel (x − a), then put x = a.
A very useful result
lim(x→a) (xⁿ − aⁿ)/(x − a) = n·aⁿ⁻¹. For example lim(x→2) (x⁵ − 32)/(x − 2) = 5 × 2⁴ = 80.
Limits of trigonometric functions
sin x and cos x are smooth, so lim(x→a) sin x = sin a and lim(x→a) cos x = cos a.
Two standard results (x in radians):
- lim(x→0) sin x / x = 1
- lim(x→0) (1 − cos x)/x = 0
Why is the first one true? On a unit circle, for a small angle x, the arc (length x) and the half-chord (length sin x) are almost the same. A squeeze argument with areas proves it: cos x ≤ sin x / x ≤ 1, and both ends go to 1.
Trick: make the angle and the bottom match. lim sin 5x / x = 5 × lim (sin 5x)/(5x) = 5.
Limits of exponential and logarithmic functions
eˣ and ln x are smooth where they are defined, so for them we can put in the value (for ln x we need x > 0).
Two standard results:
- lim(x→0) (eˣ − 1)/x = 1
- lim(x→0) ln(1 + x)/x = 1
Again match the inside with the bottom: lim (e³ˣ − 1)/x = 3 × lim (e³ˣ − 1)/(3x) = 3.
How limits lead to derivatives
The slope of a curve at one point needs a gap of zero, which we cannot divide by. So we take a small gap h and let h → 0. That limit is the derivative. You will study it in the next lesson.
Try it
Use a calculator in radian mode. Find sin(0.1)/0.1, sin(0.01)/0.01 and sin(0.001)/0.001. Predict where the values are heading. Then open the 3D free play, pick sin x / x and make h tiny to check.
Key formulas and definitions
- lim(x→a) f(x) = L when f(x) → L from both sides
- Limit exists ⇔ lim(x→a⁻) f(x) = lim(x→a⁺) f(x)
- lim(x→a) (xⁿ − aⁿ)/(x − a) = n·aⁿ⁻¹
- lim(x→0) sin x / x = 1 (x in radians)
- lim(x→0) (1 − cos x)/x = 0
- lim(x→0) (eˣ − 1)/x = 1
- lim(x→0) ln(1 + x)/x = 1
Worked examples
1. Find lim(x→3) (x² + 2x − 1).
It is a polynomial, so put x = 3: 9 + 6 − 1 = 14.
2. Find lim(x→2) (x² − 4)/(x − 2).
Putting x = 2 gives 0/0. Factorise: (x − 2)(x + 2)/(x − 2). Cancel (x − 2): x + 2. Now put x = 2: answer 4.
3. f(x) = x + 1 for x < 1 and f(x) = 3x for x ≥ 1. Does lim(x→1) f(x) exist?
Left limit: 1 + 1 = 2. Right limit: 3 × 1 = 3. They are not equal, so the limit does not exist.
4. Find lim(x→1) (x¹⁰ − 1)/(x − 1).
Use (xⁿ − aⁿ)/(x − a) → n·aⁿ⁻¹ with n = 10, a = 1: 10 × 1⁹ = 10.
5. Find lim(x→0) sin 4x / sin 2x.
Write it as [sin 4x/(4x)] × 4x ÷ ([sin 2x/(2x)] × 2x). Both brackets → 1. So the answer is 4x/2x = 2.
6. Find lim(x→0) (e²ˣ − 1)/x.
Multiply and divide by 2: 2 × (e²ˣ − 1)/(2x). The fraction → 1, so the answer is 2.
7. Find lim(x→0) ln(1 + 3x)/x.
Write as 3 × ln(1 + 3x)/(3x). The fraction → 1 as 3x → 0. Answer: 3.
8. Find lim(x→0) (1 − cos 2x)/x².
1 − cos 2x = 2 sin²x. So the limit is 2 × (sin x / x)² → 2 × 1 = 2.
Common mistakes
- Writing 0/0 as the answer. 0/0 only means 'do more work': factorise, cancel or use a standard limit.
- Using degrees in sin x / x. The result 1 is true only when x is in radians.
- Thinking the limit must equal f(a). A graph can have a hole at a and still have a limit there.
- Checking only one side. If left and right limits differ, the limit does not exist.