How to Learn Limits and Continuity Quickly and Correctly | Quick Notes

To learn Limits and Continuity quickly and correctly, begin with the idea of approaching a point. A limit asks what value a function’s outputs get close to when its inputs move close to a particular number. Continuity asks whether the function reaches that expected value without a break at the point itself. You do not need to memorise complicated language first: start by reading a small table, watching a graph and comparing the left and right sides. These eduKateSG Mathematics Quick Notes teach the basic rules with clear examples, common mistakes and short practice.

• Read a limit: lim as x approaches a of f(x) asks what f(x) approaches.
• Try substitution: for a continuous expression at a, substitute x = a.
• If you get 0/0: simplify the expression when possible, then reconsider the limit.
• Check both directions: a two-sided limit exists only when left and right limits agree as finite real values in this beginner setting.
• Test continuity: f(a) must exist, the finite limit must exist, and the two must be equal.
• Check: use nearby numbers or a sketch to confirm the conclusion.

First, imagine approaching the number 3 along a number line. You might move from 2.9 to 2.99 and then 2.999, getting closer from the left. Or you might move from 3.1 to 3.01 and then 3.001, getting closer from the right. Now let f(x) = x + 2. From the left, the output approaches 5; from the right, it also approaches 5. We write lim as x → 3 of (x + 2) = 5. In this easy example, substituting 3 gives the same answer because x + 2 is continuous there. The word “approaches” describes nearby input values, not merely the function’s value at the exact point.

Second, understand why a limit and a function value are not automatically the same. Consider f(x) = (x² − 9)/(x − 3) for x ≠ 3. Factorise the numerator: (x − 3)(x + 3). For every allowed x other than 3, the fraction equals x + 3. As x gets close to 3, the output gets close to 6. But at x = 3 the original denominator is zero, so the original function is undefined there. The limit is 6 even though f(3) does not exist. Picture a straight-line graph with one missing point or “hole” at (3, 6).

Third, test continuity using three checks. To be continuous at x = a, a function must have a defined value f(a); its finite two-sided limit as x approaches a must exist; and that limit must equal f(a). For f(x) = x² at a = 2, the function value is 4, the limit is 4 and the values match, so it is continuous. In the previous fraction example, the limit at x = 3 is 6 but f(3) does not exist, so that function is not continuous there. If we separately define f(3) = 6 while keeping the expression for x ≠ 3, the new piecewise function becomes continuous at 3.

Worked example: factorise before evaluating the limit

Find lim as x → 2 of (x² − 4)/(x − 2). Direct substitution produces 0/0, which is indeterminate, not an answer of zero. Factorise x² − 4 = (x − 2)(x + 2). For x ≠ 2, cancel the factor x − 2 to get x + 2. Now let x approach 2: the output approaches 4. The original expression is undefined at x = 2, but the limit equals 4. The cancellation is valid for nearby x ≠ 2, which is all the limit requires.

Worked example: use a small table

For f(x) = (x² − 9)/(x − 3) at nearby allowed inputs, calculate f(2.9) = 5.9 and f(2.99) = 5.99. From the other side, f(3.01) = 6.01 and f(3.1) = 6.1. The outputs approach 6 from both directions. The simplified rule x + 3 confirms the same result. A table can suggest what is happening, but the algebra explains why the values approach 6. Do not mistake a calculator error at x = 3 for proof that there is no limit.

Left-hand and right-hand limits

A left-hand limit asks what happens as x approaches a using values smaller than a. A right-hand limit uses values larger than a. Imagine the piecewise rule f(x) = x + 1 for x < 2, and f(x) = x + 2 for x ≥ 2. From the left, outputs approach 3. From the right, they approach 4. Because 3 and 4 are different, the ordinary two-sided limit at x = 2 does not exist. Even though f(2) is defined as 4, the jump between the two sides prevents continuity.

Worked example: a continuous piecewise function

Define f(x) = x + 1 when x < 2, and f(x) = 5 − x when x ≥ 2. The left-hand limit at 2 is 2 + 1 = 3. The right-hand limit is 5 − 2 = 3. The function’s actual value at x = 2 uses the second rule: f(2) = 3. All three values agree, so the function is continuous at x = 2. A function can have different expressions on either side and still be continuous if the pieces join properly.

Three kinds of breaks in a graph

• Removable hole: the two-sided limit exists, but the function is missing or assigned the wrong value at that point.
• Jump: the left and right limits approach different finite values.
• Infinite behaviour: outputs grow without bound in at least one direction, such as 1/x near zero.
• No break: the finite limit equals the actual function value, giving continuity at that point.

What does an infinite limit mean?

Consider f(x) = 1/x near x = 0. From the right, 1/0.1 = 10 and 1/0.01 = 100; outputs increase without bound. From the left, 1/(−0.1) = −10 and 1/(−0.01) = −100; they decrease without bound. Thus the right-hand behaviour approaches positive infinity and the left-hand behaviour approaches negative infinity. The two-sided finite limit at zero does not exist. Infinity is not an ordinary real-number answer you can substitute as though it were a number. It describes unbounded behaviour.

The first limit laws to use

If lim f(x) = L and lim g(x) = M as x approaches the same point, then the limit of their sum is L + M, of their product is LM, and of their quotient is L/M when M ≠ 0, under the usual finite-limit assumptions. These rules explain why substituting directly works for polynomials: addition and multiplication preserve the limiting values. Division needs care when the denominator approaches zero. The expression 0/0 signals that you need to inspect the form further; you must not cancel arbitrary terms or declare an automatic result.

Why limits matter for differentiation

A derivative measures the limit of average gradients over smaller and smaller intervals. For example, the gradient between x and x + h on y = x² is [(x + h)² − x²]/h = 2x + h when h ≠ 0. As h approaches zero, that expression approaches 2x, the instantaneous gradient. You never actually divide by h = 0; you simplify for non-zero h and then consider the limit. This is the bridge between elementary algebra, graphs and differentiation.

A quick decision routine

1. Write the input value x is approaching and name the function.
2. Check direct substitution first.
3. If the value is a valid real number, report it for a continuous expression.
4. If the result is 0/0, try factorising and cancelling an entire non-zero factor for nearby inputs.
5. If the function is piecewise, compare the left-hand and right-hand limits.
6. To decide continuity, also calculate the actual function value at the point.
7. Check with a sketch or two nearby values and state whether the limit exists.

Common mistakes and their repair

• Calling 0/0 equal to zero: it is an indeterminate form, not a final limit.
• Assuming an undefined point has no limit: neighbouring outputs may still approach one value.
• Checking only one side: a two-sided limit requires agreement between both directions.
• Confusing f(a) with the limit: evaluate them separately when continuity is in question.
• Cancelling terms across a sum: cancel only legitimate common factors after factorisation.
• Treating infinity as an ordinary finite output: describe unbounded behaviour accurately.
• Concluding continuity from a matching left and right limit alone: also check f(a) is defined and equal to the limit.

Mini-test: try four limits and a continuity check

1. Find lim as x → 4 of (2x + 1).
2. Find lim as x → 3 of (x² − 9)/(x − 3).
3. Find lim as x → 2 of (x² − 4)/(x − 2).
4. For f(x) = x + 1 if x < 2 and f(x) = x + 2 if x ≥ 2, does the two-sided limit at 2 exist?
5. A new function equals (x² − 9)/(x − 3) when x ≠ 3 and equals 6 when x = 3. Is it continuous at x = 3?

Answers with short explanations

1. Substitute x = 4: 8 + 1 = 9. 2. Simplify the fraction for x ≠ 3 to x + 3, so the limit is 6. 3. Simplify to x + 2 for x ≠ 2, giving 4. 4. No: the left-hand limit is 3 and the right-hand limit is 4. 5. Yes: the two-sided limit is 6 and the defined value at 3 is also 6.

Your ten-minute limits and continuity study routine

1. Minutes 1–2: explain “approaching a number” using values just left and right of it.
2. Minutes 3–4: evaluate two easy limits by substitution.
3. Minutes 5–6: factorise a 0/0 expression and find its limit.
4. Minutes 7–8: compare the two sides of a piecewise function.
5. Minutes 9–10: test continuity using all three conditions and explain a removable hole.

The most important distinction is between a function’s value at a point and what its values approach near that point. Once you can check both directions and explain a missing point, limits and continuity become a clear foundation for calculus.

Continue the Mathematics Quick Notes learning pathway

Build on this lesson with Differentiation · Functions and Function Notation · Quadratic Graphs · Integration. Try one different question tomorrow without copying the worked steps.

The original longer teaching reference, Secondary 1 Mathematics Tutor Clementi | Small Groups Tutorials, remains unchanged.