Limits and Continuity

The gradient of a straight line is rise over run between two points. A curve has a different gradient at every point, so that formula cannot give the gradient at one point. A limit answers that question: it asks what a calculation approaches as its input approaches a value, without calculating at that value.

What problem is calculus actually solving?

Calculus measures change on curves, where straight-line methods stop working. A straight line has one gradient everywhere. A curve steepens as you move along it, so no single number describes it.

Ask instead for the gradient at one point, and draw the tangent there.

To calculate it, find the gradient of the chord from that point to a nearby second point, then slide the second point toward the first. The value the chord gradient approaches is the gradient of the curve. See The Gradient of a Curve and Limits.

What is a limit?

A limit is the value a function approaches as its input approaches a point. What the function does at the point itself does not matter, and it need not be defined there at all.

$\frac{\left(x^2-4\right)}{\left(x-2\right)}$(x² − 4)/(x − 2) is undefined at x = 2, and its limit there is 4. Factor the numerator as (x + 2)(x − 2) and cancel the common factor. Away from x = 2 the function is x + 2, which is near 4 when x is near 2.

Does a limit have to agree from both sides?

Yes. A one-sided limit is the value a function approaches from the left only, or from the right only. The two-sided limit exists only when the left limit and the right limit are equal.

A step function shows the rule. Both one-sided limits exist at the jump, but they disagree, so the limit does not exist there. See One Sided Limits.

How do you actually calculate a limit?

Substitute first. If substitution gives a number, that number is the limit. The limit laws allow this: a limit passes through sums, products and quotients. See The Limit Laws.

Substitution fails when it gives an indeterminate form such as $\frac{0}{0}$0/0 or $\infty$∞/$\infty$∞. That does not mean there is no limit. It means the expression in this form has not decided the question.

1. Substitute. A number means you are done.
2. If you get $\frac{0}{0}$0/0, rewrite the expression: factor and cancel, rationalize a surd, or combine the fractions.
3. Substitute again into the new expression.

$\lim \frac{\left(x^2-9\right)}{\left(x-3\right)}$lim (x² − 9)/(x − 3) as $x\to 3$x → 3: factor the numerator to get $\frac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)}$(x + 3)(x − 3)/(x − 3), cancel the (x − 3), then substitute x = 3 into x + 3. The limit is 6. See Indeterminate Forms.

What happens to a function far out?

A limit at infinity asks what a function approaches as x grows without bound. For a fraction of two polynomials, only the highest power of x matters when x is large.

$\frac{\left(3x^2+5\right)}{\left(x^2-1\right)}\to 3$(3x² + 5)/(x² − 1) → 3 as $x\to \infty$x → ∞. Divide the numerator and the denominator by $x^2$x² to get $\frac{\left(3+\frac{5}{x^2}\right)}{\left(1-\frac{1}{x^2}\right)}$(3 + 5/x²)/(1 − 1/x²). As x grows, $\frac{5}{x^2}$5/x² and $\frac{1}{x^2}$1/x² both approach 0, leaving 3. The line y = 3 is a horizontal asymptote.

Compare the degrees: equal degrees give the ratio of the leading coefficients, a smaller degree on top gives 0, and a larger degree on top means the function grows without bound. See Limits at Infinity.

How do you find a limit you cannot compute directly?

Trap the function between two others whose limits you know. If the upper and lower functions approach the same value, the function between them must approach it too. This is the squeeze theorem. See The Squeeze Theorem.

The most important use is the limit of sin x over x as x approaches 0. Substituting x = 0 gives $\frac{0}{0}$0/0. Near 0 the ratio is trapped between cos x and 1, and both approach 1, so the limit is exactly 1 when x is in radians. See The Limit of sin x over x.

For small x in radians, $\sin x\approx x$sin x ≈ x, $\tan x\approx x$tan x ≈ x and $\cos x\approx 1-\frac{x^2}{2}$cos x ≈ 1 − x²/2. These are the small-angle approximations. See Small-Angle Approximations.

What does continuity mean?

A function is continuous at a point when three conditions hold: the function is defined there, the limit exists there, and the limit equals the value. Informally, you can draw the graph through that point without lifting the pencil. See Continuity at a Point.

A discontinuity is one of three types. A removable discontinuity is one missing or misplaced point: the limit exists, but the value does not equal it. At a jump discontinuity the left and right limits differ. At an infinite discontinuity the curve runs off along a vertical asymptote, so there is no limit. See Types of Discontinuity.

The intermediate value theorem says that a function continuous on an interval takes every value between its values at the two ends. So a continuous function that is negative at one end and positive at the other must equal zero somewhere between: only a jump could cross the axis without touching it, and continuity rules out the jump.

This is the guarantee behind every numerical root-finder, including the iteration methods in the applications of differentiation: a sign change between two points traps a root between them. See The Intermediate Value Theorem.

What can you do once you have a derivative?

Differentiating from first principles carries out the limit in full: write the chord gradient $\frac{\left(f\left(x+h\right)-f\left(x\right)\right)}{h}$(f(x + h) − f(x))/h, simplify, then let h approach 0. For $f\left(x\right)=x^2$f(x) = x² the chord gradient simplifies to 2x + h, so the derivative is 2x. See From First Principles.

The same working on $x^3$x³ gives $3x^2$3x², and the pattern is the rule for powers: multiply by the power, then take one off the power. So $x^5$x⁵ differentiates to $5x^4$5x⁴. See Differentiating Powers.

The derivative at a point is the gradient of the tangent there. The normal is perpendicular to the tangent, and perpendicular gradients multiply to −1, so the gradient of the normal is $-1/m$−1 / m. See Tangents and Normals.

A stationary point is where the derivative is zero. To find one, set the derivative equal to zero and solve. See Stationary Points.

What is integration, and why is it the reverse?

Integration undoes differentiation. Differentiating $x^3$x³ gives $3x^2$3x², so integrating $3x^2$3x² returns $x^3$x³. Differentiation destroys any constant term, so an integral always ends with + c. See Integration.

Integration also measures the area under a curve. Slice the region into thin rectangles and add their areas. The total improves as the rectangles get thinner. See Area Under a Curve.

The sum $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots$1/2 + 1/4 + 1/8 + … approaches 1. The sum $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots$1 + 1/2 + 1/3 + 1/4 + … grows past every bound, although its terms also shrink to 0. Shrinking terms are not enough for a sum to converge. See Two Infinite Sums, Opposite Fates.

The mistakes worth naming

• Reading $\frac{0}{0}$0/0 as "no limit". Substitution failed. Rewrite the expression and substitute again.
• Confusing the limit with the value. A function can have a limit at a point where it has no value.
• Checking one side only. The left and right limits must agree, and a jump is invisible from one side.
• Using small-angle approximations in degrees. $\sin x\approx x$sin x ≈ x holds only in radians.
• Assuming shrinking terms give a finite sum. The harmonic series diverges.

Learn this properly in the app

Every lesson linked above is in Math Challenge, with a diagram, a worked example and practice questions. The functions and graphs this guide assumes are in functions and rational functions, the differentiation rules that replace first principles are in the rules of differentiation, and convergence is covered in sequences and series.