Decide whether a function is continuous or differentiable at a point using one-sided limits, derive derivatives from first principles, and find and prove patterns in higher derivatives.
Each is in the Maths AA glossary, with every other term in the subject.
A function is continuous at if is defined,
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A point where a continuous graph changes direction sharply, so the gradients from the left and the right are different (a corner) or tend to (a cusp, as for at ). The function is not differentiable there.
A function is differentiable at if exists, so its graph has a single, non-vertical tangent there. A differentiable function is continuous, but not every continuous function is differentiable.
Finding a derivative directly from its definition , by simplifying the gradient of a chord and then letting tend to .
A point where a function is not continuous, such as a jump (the one-sided limits differ), a hole (the limit exists but is not ) or a vertical asymptote.
A derivative found by differentiating more than once. The th derivative of is written or .
The value approaches as tends to from one side only: from values less than , and from values greater than .
A function defined by different rules on different parts of its domain, for example for and for .
The derivative was introduced as the limit of the gradient of a chord (5.1). That definition leaves two matters open: when the limit actually exists, and how it can be used directly to find derivatives. Both depend on a closer look at limits, and at what it means for a graph to be unbroken (continuous) and smooth (differentiable) at a point.
These ideas explain why the differentiation rules work, and also where they stop working. A function can be continuous at a point without having a derivative there, as at the sharp corner of , and functions defined in pieces, which model situations such as tax bands and parking charges, can fail to be continuous or differentiable at the points where the pieces join.
The statement means that can be made as close to as desired by taking
Sometimes approaches different values depending on whether approaches from below or from above. A one-sided limit looks at only one of these directions.
For , every negative gives
A limit converges if it equals a finite number . Otherwise it diverges. A limit can diverge in three different ways: the values can grow without bound (towards or ), the two one-sided limits can disagree, or the values can oscillate without settling down. The same words describe sequences and series (1.8), whose sums converge or diverge in the same sense.
Limit | Behaviour | Converges? |
|---|---|---|
Writing is a convenient way of describing how the limit diverges, but
Informally, a function is continuous at if its graph can be drawn through without lifting the pen. The formal version makes this precise using limits.
Polynomials, , and are continuous for every real . Rational functions, and
A discontinuity is a point where at least one of the three conditions fails. Discontinuities are usually described by how the graph breaks.

Type | What fails | Example at |
|---|---|---|
Removable (a hole) | The limit exists, but is undefined or not equal to it |
A removable discontinuity is named that way because redefining a single value would remove it: setting for
A piecewise function uses different rules on different parts of its domain. Each piece is usually a continuous function on its own, so the only places where a discontinuity can appear are the joins. At a join , compare three numbers: the left-hand limit (from the left piece), the right-hand limit (from the right piece) and (from whichever piece includes ). The function is continuous at when all three are equal.
Real pricing rules are often piecewise, and their jumps have a practical meaning. A car park that charges for each started hour has a cost function that is constant within each hour and jumps at every whole number of hours.
When a piecewise function contains an unknown constant, continuity at the join gives an equation for it: set the left-hand limit equal to the right-hand limit.
The derivative of at is the limit of the gradient of the chord from to a nearby point
Like any limit, it may or may not exist. The quotient is called a difference quotient; taking uses chords to the right of
A function is differentiable on an interval if it is differentiable at every point of it. Polynomials, , and are differentiable everywhere, and is differentiable for
The graph of is a V shape, with a sharp corner at the origin. To the left of the graph is the line , with gradient ; to the right it is
A cusp is a sharper kind of corner, where the gradients on the two sides tend to and . The function
If is differentiable at , it must also be continuous there. For ,
As , the fraction tends to the finite number and tends to , so the product tends to
Function | Continuous at ? | Differentiable at ? | Reason |
|---|---|---|---|
yes | yes | smooth, |
For a piecewise function whose pieces are each differentiable, two conditions at a join make it differentiable there:
Continuity: the pieces meet, so their values at are equal.
Matching gradients: the derivatives of the two pieces are equal at , so the join is smooth.
The first condition cannot be skipped. Matching gradients alone are not enough, because a function with a jump is never differentiable.
Replacing the fixed point by a general turns the derivative at a point into the derivative function. Finding a derivative directly from this limit is called differentiation from first principles.

The difference quotient cannot simply be evaluated at , because that gives . The method is to simplify it algebraically until the factor in the denominator cancels, and only then let
Write down , replacing every in by .
Simplify the numerator
For , the binomial expansion of does the same job. Every term except the first contains , and after dividing by every term except one still contains
The same argument works for any positive integer . By the binomial theorem (1.9),
Subtracting and dividing by leaves plus terms that all contain , so the limit is
First principles also confirm the rules for constant multiples and sums: in the difference quotient, a constant factor stays outside, a constant term cancels in , and the terms of a sum can be handled separately.
For a reciprocal, the two fractions in the numerator are combined over a common denominator before dividing by .
A square root is handled by rationalising the numerator, the same technique as for the limit
The derivative of can also be found from first principles, using the compound angle identity (3.10) and two standard limits for
The first of these is studied in 5.13; here both are taken as known. Then
As , this tends to . The two standard limits are only true when
Differentiating the second derivative (5.7) gives the third derivative, and so on. Any derivative found by differentiating more than once is a higher derivative.
The brackets in distinguish a derivative from a power. Without them, could be confused with the fourth power
For many functions the higher derivatives follow a pattern, which can be written as a formula for . Working out the first three or four derivatives, without simplifying the numbers too far, usually makes the pattern visible.
First derivatives | ||
|---|---|---|
Alternating signs are produced by , and growing products such as are factorials. For
A pattern found from the first few derivatives is only a conjecture. Proof by induction (1.15) turns it into a theorem: show the formula for , then show that if it holds for , differentiating once more gives the formula for .
These topics appear mainly on the non-calculator HL paper, often as a short "prove from first principles" question or as a piecewise function with unknown constants. Induction on higher derivatives is a common longer question, and the ideas of continuity and differentiability underpin the investigations on Paper 3.
A longer question often combines higher derivatives with induction. The early parts are routine differentiation, the conjecture must be written as a formula in , and the proof uses the chain rule in the inductive step.
I can find left-hand and right-hand limits and decide whether a limit exists.
I can decide whether a limit converges or diverges, and describe how it diverges.
I can state the three conditions for continuity at a point and classify a discontinuity as removable, jump or infinite.
I can check a piecewise function for continuity at a join, and find a constant that makes it continuous.
I can explain why is continuous but not differentiable at , and recognise corners, cusps and vertical tangents.
I can find constants that make a piecewise function differentiable, using continuity and matching gradients.
I can derive the derivatives of
21 exam-style questions on this subtopic, written in the style of the IB papers. The first has its markscheme open. Counted on
Let , for . The th derivative of is denoted by .
Find and .
2 marks
Let , for . The
Let , where
One of 39 short multiple-choice questions on 5.12. Pick an answer to see why it's right or wrong.
Which of these is not one of the conditions for to be continuous at ?
The entries of the mathematics: analysis and approaches formula booklet that 5.12 uses. Each opens in the explained booklet.
Substituting gives , which has no value, so is not defined. For , factorise and cancel:
As approaches , approaches , so
Find and , and hence state whether exists.
For , , so the left-hand limit is .
For , , so the right-hand limit is
The one-sided limits are different (), so does not exist.
approaches a finite value |
yes, to |
grows without bound on both sides () | no |
from the left, from the right | no |
approaches as grows | yes, to |
oscillates between and for ever | no |
(b)
(c)
(a) Substituting gives , so rewrite the expression. Multiply the numerator and denominator by (rationalising the numerator):
Now let : the limit converges to
(b) As from either side, is a small positive number, so
(c) As increases, keeps taking every value between and and never settles towards one value. The limit diverges by oscillation.
.
A function is continuous on an interval if it is continuous at every point of the interval.
Jump | The left-hand and right-hand limits are finite but different | at |
Infinite | on at least one side | at |
Left-hand limit, from : .
Right-hand limit, from :
Value: belongs to the second piece, so .
All three equal , so is continuous at .
(b) State the values of at which is not continuous, and describe the type of discontinuity.
(a) Just under 1 hour costs USD 3, so . Just over 1 hour costs USD 5, so . A stay of exactly 1 hour is "up to 1 hour", so .
(b) The one-sided limits differ at and at , so has a jump discontinuity at each of them. It is continuous at every other in
Each piece is a polynomial, so is continuous everywhere except possibly at .
Left-hand limit and value: . Right-hand limit: .
For continuity these must be equal:
Check: with , both pieces give at .
Differentiability. The difference quotient at is
For this equals , and for it equals . So the right-hand limit is and the left-hand limit is . They are different, so does not exist and is not differentiable at .
yes | no | corner: one-sided gradients and |
yes | no | cusp |
yes | no | vertical tangent |
(with ) | no | no | jump |
(b)
(a) Continuity was shown earlier: both pieces give at . Gradients: the left piece has derivative , which is at ; the right piece has derivative . They match, so is differentiable at , with . The line is the tangent to at , so the pieces join smoothly.
(b) Both pieces give at , so is continuous there. The gradients are from the left and from the right, which are different. The graph has a corner at
Differentiable implies continuous, so the pieces meet at :
The gradients match at . The derivative of is , which is at ; the derivative of is :
Substitute (2) into (1): , so and .
Check: at both pieces give , and both have gradient .
This definition is in the formula booklet. Here is the horizontal distance between the two points of the chord, and is the vertical distance.
Divide by , cancelling the common factor.
Let in what remains.
Write . Then
The cancellation of in the third line is valid because while the limit is being taken.
Expand: . Then
Subtract ; the terms ,
Divide by and take the limit:
Combine the fractions in the numerator:
Divide by :
Let :
This agrees with the power rule: .
(b) Use first principles to find the gradient of at the point where .
(a) Multiply the numerator and denominator by :
Let :
which agrees with
(b) At , and
Differentiate repeatedly:
Each differentiation lowers the degree of a polynomial by one. A polynomial of degree becomes a constant after four differentiations, and every derivative after that is .
, , |
, , , | repeats every 4; |
, , |
, , |
, , |
Let be the statement .
Base case. For , the product rule gives
Inductive step. Assume is true for some :
This is .
Conclusion. is true, and true implies true. Therefore, by the principle of mathematical induction, is true for all
(b) Hence find the equation of the tangent to the graph of at the point where . [3 marks]
(a) Expand :
Subtract :
Then
The marks are for: the correct , the definition with written, the simplified numerator, cancelling , and taking the limit to reach the given answer.
(b) and
where and are constants.
(a) Given that is differentiable at , find the values of and . [5 marks]
(b) Using these values, determine whether exists. Justify your answer. [3 marks]
(a) For continuity at : the left piece gives , and the right-hand limit is . So .
For matching gradients: the left piece has derivative , which is at ; the right piece has derivative , which tends to
(b) For , , which tends to as
(a) Find , and . [3 marks]
(b) Write down a conjecture for , for . [1 mark]
(c) Prove your conjecture by mathematical induction. [6 marks]
(a) Write and use the chain rule; each derivative brings down the power and multiplies by :
(b) The coefficients are , and
(c) Let be the statement
Base case:
Assume is true for some , so
This is , since .
is true, and true implies true, so by mathematical induction is true for all
I can find higher derivatives, conjecture a formula for and prove it by induction.
A1
Prove by mathematical induction that
6 marks
consider :
Show that .
3 marks
Prove by mathematical induction that
Hence find the first four non-zero terms of the Maclaurin series for .
4 marks
Hence find
Find, in terms of , the coefficient of in the Maclaurin series for , where
Show that .
2 marks
Hence, using the definition of the derivative from first principles, find .
3 marks
Note: Award A0 if is not used or implied.
assume true for , so for some M1
consider : M1
A1
A1
since true for , and true for implies true for , the result is true for all by mathematical induction R1
Note: Award the final R1 only if at least four of the previous marks have been awarded.
7 marks
2 marks
3 marks