Limits and Continuity

Updated 7 October 2026

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 that does not happen at a steady rate, where rise over run stops working. A straight line has one gradient everywhere, but a curve such as $y=x^2$y = x² steepens as you move along it, so no single number describes it. Calculus asks for the gradient at one point, the steepness of the tangent there. A chord gives an estimate: from (1, 1) to (3, 9) on $y=x^2$y = x², the gradient is 8 ÷ 2 = 4. Moving the second point closer improves the estimate, and the value it approaches is the gradient at x = 1. That value is a limit, which is why limits come first. Area is the second question: a region with a curved edge has no area formula, and integration answers it with the same idea of ever closer approximations. Check the chord: the rise is 9 − 1 = 8 and the run is 3 − 1 = 2.

Drawing the tangent by eye gives only a rough value.

With the second point at (2, 4) the chord gradient is 3, and with $\left(\frac{3}{2},\frac{9}{4}\right)$(3/2, 9/4) it is $\frac{5}{2}$5/2, closing in on 2. 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. Saying that f(x) has limit L as $x\to c$x → c means f(x) can be made as close to L as you like by taking x close enough to c. For f(x) = 2x + 1, as x approaches 3 the outputs approach 7: at x = 3 itself, 2 × 3 + 1 = 7. The value at c is ignored, so f need not be defined there. The function $\frac{\left(x^2-1\right)}{\left(x-1\right)}$(x² − 1)/(x − 1) gives $\frac{0}{0}$0/0 at x = 1, but everywhere else it equals x + 1, so its limit there is 2. A limit can also fail to exist, when the function jumps or grows without bound near c. Check with inputs on both sides: a little either side of x = 1, x + 1 is just above or below 2, and 1 + 1 = 2 is what both sides approach.

$\frac{\left(x^2-4\right)}{\left(x-2\right)}$(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 limit exists only when the function approaches the same value from both sides. The left-hand limit, $x\to c^{-}$x → c⁻, uses inputs below c only, and the right-hand limit, $x\to c^{+}$x → c⁺, uses inputs above c only. Take f(x) = x + 1 for x < 2 and f(x) = x + 2 for $x\ge 2$x ≥ 2. From the left the outputs approach 2 + 1 = 3, and from the right they approach 2 + 2 = 4, so the limit at 2 does not exist, even though f(2) = 4. Both sides are required because the definition asks f(x) to be close to one number for every x near c, and here the two sides stay near different numbers. The steps change at the end of a domain: $\sqrt{x}$√x has no inputs below 0, so only the right-hand limit is used there. Check the gap from the two values: 4 − 3 = 1, the height of the jump.

Every step function behaves this way at each of its steps. See One Sided Limits.

How do you actually calculate a limit?

Substitute first: if putting x = c into the expression gives a number, that number is the limit. For $x^2+3x$x² + 3x as x approaches 2, substitution gives 4 + 6 = 10, so the limit is 10. See The Limit Laws. The limit laws allow this: the limit of a sum is the sum of the limits, and the same holds for products, and for quotients whose denominator does not tend to 0. The steps change when substitution gives $\frac{0}{0}$0/0, which says the expression must be rewritten, not that the limit is missing. For $\frac{\left(x^2-16\right)}{\left(x-4\right)}$(x² − 16)/(x − 4) at x = 4, factor the top as (x + 4)(x − 4), cancel, and substitute into x + 4 to get 8. A nonzero number over 0, such as $\frac{1}{0}$1/0, signals that the function grows without bound. Check with an input close to c: at an input just above 2, $x^2+3x$x² + 3x gives just over 10.

The same holds for $\infty$∞/$\infty$∞, the other indeterminate form.

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 is the value a function approaches as x grows without bound. For a fraction of two polynomials, only the highest powers matter when x is large, so divide every term by the highest power of x in the denominator. For $\frac{\left(4x+1\right)}{\left(2x-3\right)}$(4x + 1)/(2x − 3), dividing by x gives $\frac{\left(4+\frac{1}{x}\right)}{\left(2-\frac{3}{x}\right)}$(4 + 1/x)/(2 − 3/x). As x grows, $\frac{1}{x}$1/x and $\frac{3}{x}$3/x shrink to 0, leaving $\frac{4}{2}=2$4/2 = 2, so the limit is 2 and y = 2 is a horizontal asymptote. The lower terms fade because they grow more slowly: at x = 1,000, the 4x term is 4,000 and the 1 adds only one part in 4,000. When the numerator has the higher degree, there is no finite limit, and when the denominator does, the limit is 0. Check with a large input: at x = 1,000 the fraction is 4,001 ÷ 1,997, just above 2, since 2 × 1,997 = 3,994.

$\frac{\left(3x^2+5\right)}{\left(x^2-1\right)}\to 3$(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.

In short, compare the degrees. See Limits at Infinity.

How do you find a limit you cannot compute directly?

When a limit cannot be computed directly, trap the function between two simpler ones whose limits are known. If $g\left(x\right)\le f\left(x\right)\le h\left(x\right)$g(x) ≤ f(x) ≤ h(x) near c, and g and h both approach L, then f must approach L too: this is the squeeze theorem. See The Squeeze Theorem. Take $f\left(x\right)=x^2\sin \left(\frac{1}{x}\right)$f(x) = x² sin(1/x) as x approaches 0. Substitution fails, and $\sin \left(\frac{1}{x}\right)$sin(1/x) swings between −1 and 1 forever, but that gives $-x^2\le f\left(x\right)\le x^2$−x² ≤ f(x) ≤ x². Both outer functions approach 0, so f(x) approaches 0 as well. The theorem holds because f has nowhere else to go: every value lies between two numbers that close in on L together. The bound needs care. It must hold for all x near c, and both sides must approach the same limit, since bounds with different limits prove nothing. Check with a number: at $x=\frac{1}{10}$x = 1/10, the bounds are $\pm \frac{1}{100}$±1/100, since $\frac{1}{10}\times \frac{1}{10}=\frac{1}{100}$1/10 × 1/10 = 1/100, and f lies between them.

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 f is continuous at x = c when three conditions hold: f(c) is defined, the limit as x approaches c exists, and the two are equal. Informally, the graph can be drawn through that point without lifting the pencil. See Continuity at a Point. Test $f\left(x\right)=x^2-1$f(x) = x² − 1 at x = 2: f(2) is 4 − 1 = 3, and the limit from either side is also 3. All three are needed because a value and a limit can both exist and still differ, as when one point is lifted off a curve. A piecewise function needs the test at each join. If g(x) = 2x for x < 1 and g(x) = x + k for $x\ge 1$x ≥ 1, continuity at 1 requires 2 × 1 = 1 + k, so k = 1. Check the join: the left gives 2 × 1 = 2 and the right gives 1 + 1 = 2.

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?

A derivative gives the gradient of a curve at every point, so it measures steepness and rates of change. It comes from first principles: write the chord gradient $\frac{\left(f\left(x+h\right)-f\left(x\right)\right)}{h}$(f(x + h) − f(x))/h, simplify, and let h approach 0. For $f\left(x\right)=x^2$f(x) = x², the chord gradient is 2x + h, so the derivative is 2x. See From First Principles. The gradient at x = 3 is then 2 × 3 = 6, and the tangent there passes through (3, 9) with gradient 6, so its equation is y = 6x − 9. The derivative also finds turning points, where it is zero, and rates such as speed. It works because the limit removes the chord's error term, h, leaving the exact gradient. A zero derivative can mark a maximum, a minimum or a point of inflection, so test the gradient on each side. Check the tangent at x = 3: 6 × 3 − 9 = 9, the height of the curve.

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 normal is perpendicular to the tangent, and perpendicular gradients multiply to −1, so the gradient of the normal is −1 / m. See Tangents and Normals.

To find a stationary point, set the derivative equal to zero and solve. See Stationary Points.

What is integration, and why is it the reverse?

Integration is the reverse of differentiation: to integrate a function, find a function whose derivative it is. Differentiating $x^3$x³ gives $3x^2$3x², so integrating $3x^2$3x² gives $x^3+c$x³ + c. See Integration. The constant c appears because differentiation removes constant terms, so $x^3+5$x³ + 5 and $x^3-2$x³ − 2 have the same derivative. For powers, add one to the power and divide by the new power: $x^4$x⁴ integrates to $\frac{x^5}{5}+c$x⁵/5 + c. Integration also measures area, and the reverse is why: as the right edge of the region under a curve moves, the area grows at a rate equal to the curve's height there. So a definite integral is the antiderivative at the upper end minus its value at the lower end, with no c. The area under $y=3x^2$y = 3x² from 0 to 2 is $2^3-0^3=8$2³ − 0³ = 8. The power rule fails for $x^{-1}$x⁻¹, whose integral is ln |x| + c. Check by differentiating: $\frac{x^5}{5}$x⁵/5 gives $\frac{5x^4}{5}$5x⁴/5, which is $x^4$x⁴.

Slice the region under a curve 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.

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.