Limits and Continuity
3–4 weeks · late August into September · 5–10% of the multiple-choice sectionThe problem that forces the whole subject
You want instantaneous velocity. Average velocity over an interval is easy — distance over time. Over the interval from a to a+h:
Now shrink the interval to zero to make it instantaneous. Set h = 0 and you get 0/0.
That is not a hard number to compute. It's a meaningless expression. 0/0 could be anything: 0/0 = 5 would require 5 × 0 = 0, which is true; but so would 0/0 = 7. Every answer works, so no answer is determined.
This is the crisis, and the entire limits unit is the workaround.
The dodge: don't set h = 0. Ask what the quotient approaches as h gets small. That sounds like a distinction without a difference. It isn't, and the difference is the whole concept of a limit.
What a limit actually says
Informally: limx→a f(x) = L means f(x) gets arbitrarily close to L as x gets close to a.
"Gets close to" is motion language, and there's no motion here. A function is a static object; x isn't sliding anywhere. For 190 years everyone waved at this and it mostly worked — until it didn't, and mathematicians found functions where intuition gave flatly wrong answers.
Weierstrass's fix in the 1860s replaces motion with a challenge-and-response game:
such that whenever 0 < |x − a| < δ, we have |f(x) − L| < ε.
Read it adversarially. Someone challenges you: "get f within 0.001 of L." You must produce a radius around a small enough to guarantee it. If you can meet every challenge, the limit exists.
The load-bearing detail is the leading 0 <. It explicitly excludes x = a. That single symbol is what makes limits able to talk about 0/0 at all.
One level deeper: this only works because the reals are complete — they have no gaps. A sequence of fractions can march toward √2, which is not a fraction. The real numbers were constructed (Dedekind, 1872) precisely so that anything that looks like it's approaching something actually has something to approach.
Completeness is the hidden foundation under the three big existence theorems of the course — Intermediate Value, Extreme Value, and Mean Value. All three are false over the rationals alone. BC states all three and proves none, so this sentence is the one the textbook is missing.
One-sided limits
limx→a− approaches from the left (smaller x); limx→a+ from the right.
The two-sided limit exists if and only if both one-sided limits exist and are equal. That's the definition, and it's the whole test for whether a piecewise function has a limit at its seam.
Let f(x) = x² + 1 for x < 2, and f(x) = 3x − 4 for x ≥ 2. Does limx→2 f(x) exist?
| limx→2− f(x) = 2² + 1 = 5 | from the left, x < 2, so use the first formula |
| limx→2+ f(x) = 3(2) − 4 = 2 | from the right, use the second |
| 5 ≠ 2, so the limit does not exist | a jump discontinuity at x = 2 |
Note we substituted x = 2 into a formula that officially only applies for x < 2. That's legitimate: the one-sided limit asks what the left-hand formula is heading toward at 2, and since x² + 1 is continuous, that's just its value there.
Let f(x) = x² + 1 for x < 2, and f(x) = 3x + k for x ≥ 2. Find k making f continuous at 2.
| need: left limit = right limit = f(2) | the definition of continuity, all three parts |
| left: 2² + 1 = 5 | |
| right: 3(2) + k = 6 + k | this also equals f(2), since x ≥ 2 uses this formula |
| 6 + k = 5 | set them equal |
| k = −1 |
Two-unknown versions come later. If a problem gives f(x) = x² + 1 for x < 2 and ax + b for x ≥ 2 and asks for both constants, one equation isn't enough. The second condition is differentiability — the two pieces must also have matching slopes at the seam — which requires derivatives and so lands in Unit 2 or 3.
For the record, so you recognize it: match values (4+1 = 2a+b) and match derivatives (2x at x=2 is 4, so a = 4), giving a = 4, b = −3. Full treatment in Unit 2.
The decision procedure
Given any limit, work this order. It resolves essentially every problem in the unit.
| What you get on substituting | What it means / what to do |
|---|---|
| A number | Done. Legitimate whenever f is continuous at a — all polynomials, and rationals, roots, trig, exp, log on their domains. |
| nonzero / 0 | Infinite limit. Not indeterminate. Determine the sign from each side separately; this is a vertical asymptote. |
| 0 / 0 | Indeterminate. There is a hidden common factor. Go to the toolkit below. |
| x → ±∞ | Growth-rate toolkit. This is the horizontal-asymptote question. |
| ∞−∞, 0·∞, 1∞, 00, ∞0 | Rewrite algebraically into 0/0 or ∞/∞, then L'Hôpital. |
It means the form alone doesn't determine the answer. Consider three limits as x→0, all of the form 0/0:
Same form, three different answers. The form tells you nothing; you have to dig into the specific functions. Contrast 5/0, which is not indeterminate — it always blows up. Knowing the difference tells you whether there's work to do.
Toolkit for 0/0
All four techniques do the same thing: expose and cancel the hidden factor of (x − a).
Apply the Factor Theorem to both halves of the fraction. If substituting x = a gives 0/0, then the numerator is zero at a and the denominator is zero at a. So (x − a) divides both. It's there by guarantee, not by luck. Your only job is to find it.
That's why "cancel and re-substitute" always works for rational functions, and why it's the first thing to try.
1 · Factor and cancel
Three, and they cover almost everything:
- Difference of squares: a² − b² = (a−b)(a+b). So x² − 9 = (x−3)(x+3), and x² − 5 = (x−√5)(x+√5).
- Difference of cubes: a³ − b³ = (a−b)(a² + ab + b²). Sum of cubes flips the two inner signs: a³ + b³ = (a+b)(a² − ab + b²).
- Quadratics: for x² + bx + c, find two numbers that multiply to c and add to b. For x² − 5x + 6: (−2)(−3) = 6 and (−2)+(−3) = −5, so it's (x−2)(x−3).
If the leading coefficient isn't 1 — say 2x² + 5x − 3 — either use the quadratic formula to find the roots r₁, r₂ and write 2(x−r₁)(x−r₂), or factor by grouping. The quadratic formula never fails, so when in doubt use it: x = [−b ± √(b²−4ac)] / 2a.
| limx→3 (x² − 9)/(x − 3) | substituting gives 0/0, so (x−3) must divide both |
| = limx→3 (x−3)(x+3)/(x−3) | difference of squares on top |
| = limx→3 (x + 3) | cancel — legal because x ≠ 3, only near 3 |
| = 6 | now continuous, so substitute |
The cancellation step is where the limit concept earns its keep. You are allowed to divide by (x−3) precisely because the limit never evaluates at x = 3.
2 · Rationalize — whenever there's a square root
The conjugate of √a − b is √a + b — same terms, flipped middle sign. Multiplying a pair of conjugates is difference-of-squares run forward, and squaring kills the root:
Example: (√x − 2)(√x + 2) = x − 4. No root left.
You can't just change the expression, so you multiply top and bottom by the conjugate — which is multiplying by 1. The root moves from the numerator to the denominator, where it stops causing trouble.
| limx→0 (√(x+4) − 2)/x | 0/0; there's a root, so rationalize |
| = lim (√(x+4)−2)(√(x+4)+2) / [x(√(x+4)+2)] | multiply top and bottom by the conjugate |
| = lim (x + 4 − 4) / [x(√(x+4)+2)] | numerator collapses by difference of squares |
| = lim x / [x(√(x+4)+2)] | the hidden factor of x is now visible |
| = lim 1/(√(x+4) + 2) | cancel |
| = 1/4 | substitute: 1/(2+2) |
3 · Combine complex fractions
Two facts:
- Dividing by something is multiplying by its reciprocal: (A/B) / C = A/(BC).
- To combine A/B − C/D, cross onto a common denominator: (AD − CB)/(BD).
Example: 1/5 − 1/3 = (3 − 5)/15 = −2/15.
Strategy: always clean up the numerator into a single fraction first, then deal with the outer division.
| limh→0 [ 1/(2+h) − 1/2 ] / h | 0/0; the numerator is a difference of fractions |
| numerator = [2 − (2+h)] / [2(2+h)] | common denominator 2(2+h) |
| = −h / [2(2+h)] | simplify the top |
| whole thing = −h / [2h(2+h)] | dividing by h multiplies the denominator by h |
| = −1 / [2(2+h)] | cancel h |
| = −1/4 | substitute h = 0 |
4 · Recognize a difference quotient in disguise
That last example is exactly the definition of the derivative of f(x) = 1/x at x = 2. Since f′(x) = −1/x², the answer is −1/4 in one line instead of six.
The tell: a limit as h→0 with the shape [f(something + h) − f(something)] / h. AP problems plant these deliberately — you're being tested on whether you recognize the definition, not on algebra.
This shortcut needs derivative rules from Unit 2. During the limits unit she's expected to grind the algebra, and that practice is the point. Recognize the pattern now; use it in October.
Find limh→0 [ (3+h)⁴ − 81 ] / h.
| the brute-force route: expand (3+h)⁴ | binomial expansion, four terms of mess, then cancel the 81 and divide by h |
| instead: note 81 = 3⁴ | so the expression is [f(3+h) − f(3)]/h with f(x) = x⁴ |
| that limit is f′(3), by definition | this is the derivative definition, verbatim |
| f′(x) = 4x³ | power rule (Unit 2) |
| f′(3) = 4·27 = 108 |
How to spot it: the limit is as h→0; there's a "+h" tucked inside a function; and subtracting the same function without the h. The number being subtracted (81) is the tell — it's f evaluated at the base point.
The algebra cancels to give limx→3 (x²−9)/(x−3) = 6. Confirm it without redoing the algebra: put x = 3.001 into the original expression. (3.001² − 9)/0.001 = 6.001 — a hair over 6, which is exactly what approaching a limit of 6 from the right should look like.
This works on any 0/0 you have just cancelled. It takes ten seconds and it catches a dropped sign immediately.
The special limits — each one derived
These are the limits that can't be done by algebra, because the functions involved aren't polynomials and there's no factor to cancel. Each one has to be established once, from a picture or a definition, and then it becomes a tool. Every trig and exponential derivative in the course rests on one of them.
limx→0 (sin x)/x = 1
This is the foundational one. Everything trigonometric depends on it.
An angle in radians is the arc length it cuts on a circle of radius 1. Full circle = 2π ≈ 6.283 rad = 360°. So π rad = 180°; convert by multiplying by 180/π or π/180. Common values: π/6 = 30°, π/4 = 45°, π/3 = 60°, π/2 = 90°.
Area of a circular sector of radius r and angle θ: the sector is the fraction θ/(2π) of the whole circle, so its area is (θ/2π)·πr² = ½r²θ. Clean — and clean only in radians.
Area of a triangle = ½ · base · height.
| ½ sin x ≤ ½ x ≤ ½ tan x | the three nested areas, from the picture |
| sin x ≤ x ≤ tan x | multiply through by 2 |
| 1 ≤ x/sin x ≤ 1/cos x | divide by sin x (positive, so inequalities hold); tan x/sin x = 1/cos x |
| cos x ≤ (sin x)/x ≤ 1 | take reciprocals — which flips the inequalities |
| → 1 ≤ lim (sin x)/x ≤ 1, so it's 1 | as x→0, cos x → 1; squeezed from both sides |
Find limx→0 x²·sin(1/x). Direct substitution fails badly: sin(1/x) oscillates infinitely fast as x→0 and has no limit at all.
| −1 ≤ sin(1/x) ≤ 1 | true for every input, no matter how wild — sine never leaves [−1,1] |
| −x² ≤ x²·sin(1/x) ≤ x² | multiply through by x², which is positive so the inequalities hold |
| lim(−x²) = 0 and lim(x²) = 0 | both bounds are easy, and they agree |
| lim x²·sin(1/x) = 0 | squeezed |
Note what happened: the function genuinely has no nice behavior — it wiggles infinitely often in any interval around 0. But the wiggles are being crushed by the x² envelope. The Squeeze Theorem is exactly the tool for "I can't analyze this function, but I can bound it."
Every step of that derivation used sector area = ½r²θ, which is only true in radians. In degrees the sector area picks up a factor of π/180, and the limit comes out to π/180 ≈ 0.01745 instead of 1.
Consequence: in degrees, d/dx[sin x] = (π/180)·cos x, and that ugly constant would propagate through every trig formula in mathematics forever. Radians are the unit that makes calculus clean — that's the entire reason they exist.
limx→0 (1 − cos x)/x = 0
| (1 − cos x)/x | 0/0; multiply by the conjugate (1 + cos x) |
| = (1 − cos²x) / [x(1 + cos x)] | difference of squares on top |
| = sin²x / [x(1 + cos x)] | since sin² + cos² = 1, we have 1 − cos²x = sin²x |
| = (sin x / x) · (sin x / (1 + cos x)) | split deliberately to expose the known limit |
| → 1 · (0/2) = 0 | first factor → 1; second → 0/(1+1) |
limx→0 (1 − cos x)/x² = 1/2
| (1 − cos x)/x² = sin²x / [x²(1 + cos x)] | identical first three steps as above |
| = (sin x/x) · (sin x/x) · 1/(1 + cos x) | split the x² between two copies |
| → 1 · 1 · 1/2 = 1/2 | cos 0 = 1, so the last factor is 1/2 |
Worth noticing: 1 − cos x behaves like x²/2 near zero. That's the second-order Taylor term for cosine, showing up eight months early.
limx→0 (tan x)/x = 1
| tan x / x = (sin x / cos x) / x | definition of tangent |
| = (sin x / x) · (1/cos x) | regroup |
| → 1 · 1 = 1 | cos 0 = 1 |
limx→0 (ex − 1)/x = 1
This limit is the derivative of ex at x = 0 — write out the difference quotient with a = 0 and you get exactly this expression.
So the question "why is it 1?" is really the question "what is e?" And the answer: e is defined as the base for which the exponential curve has slope exactly 1 where it crosses the y-axis. Every exponential y = bx passes through (0,1). They differ in how steeply. For b = 2 the slope there is about 0.693; for b = 3 it's about 1.099. Somewhere between 2 and 3 there's a base where the slope is exactly 1. That base is e ≈ 2.71828.
Everything convenient about e follows from that one choice — including d/dx[ex] = ex, the only function that is its own derivative.
The general version, for any base: since bx = ex·ln b, the slope at 0 is ln b. That's where the ln b in d/dx[bx] = bx·ln b comes from — check it: ln 2 ≈ 0.693 and ln 3 ≈ 1.099, matching the picture.
limn→∞ (1 + x/n)n = ex
Invest $1 at 100% annual interest. Compounded once: $2. Compounded twice a year at 50% each: (1 + ½)² = $2.25. Quarterly: (1 + ¼)⁴ ≈ $2.44. Daily: ≈ $2.7146. Continuously — the limit as n→∞ — you get e ≈ 2.71828, and no more. Compounding infinitely often doesn't give you infinite money.
Jacob Bernoulli found this in 1683 studying exactly this question. The general form with x in place of 1 is the same statement at interest rate x.
This matters later: it's why continuous growth models use ekt, and it reappears in Unit 7 as the solution to y′ = ky.
Using them: force the pattern to match
The numerator's angle and the denominator must be identical. Manufacture that, then compensate.
| limx→0 (sin 5x)/(3x) | angles don't match — 5x on top, 3x below |
| = lim (5/3) · (sin 5x)/(5x) | multiply and divide by 5 to build 5x underneath |
| = (5/3)·1 = 5/3 | the bracket is the standard limit with u = 5x |
Limits at infinity
Different question, same word. Here x runs off without bound and you're asking what f settles toward — the horizontal asymptote question in different clothing.
Rational functions: divide by the highest power in the denominator
| limx→∞ (3x² − x)/(5x² + 7) | ∞/∞; the highest denominator power is x² |
| = lim (3 − 1/x)/(5 + 7/x²) | divide every term, top and bottom, by x² |
| = 3/5 | every c/xk term → 0 |
Once you trust it, shortcut by comparing degrees:
- Top degree > bottom → ±∞. No horizontal asymptote. (If the degree exceeds by exactly 1, there's a slant asymptote instead.)
- Top degree < bottom → 0. Horizontal asymptote y = 0.
- Equal → ratio of the leading coefficients.
Same procedure as long division with numbers. Divide the leading term of what's left by the leading term of the divisor, multiply back, subtract, repeat until the remainder has lower degree than the divisor.
Example: (x² + 3x + 5) ÷ (x + 1).
| x² ÷ x = x | first term of the quotient |
| x·(x+1) = x² + x; subtract → 2x + 5 | multiply back and subtract |
| 2x ÷ x = 2 | next term of the quotient |
| 2·(x+1) = 2x + 2; subtract → 3 | remainder, degree 0, so stop |
| = x + 2 + 3/(x+1) | quotient plus remainder-over-divisor |
Find the slant asymptote of f(x) = (x² + 3x + 5)/(x + 1).
| degree 2 over degree 1 | exceeds by exactly 1 → slant asymptote exists |
| f(x) = x + 2 + 3/(x+1) | from the long division above |
| as x→±∞, the term 3/(x+1) → 0 | bottom-heavy fraction dies |
| slant asymptote: y = x + 2 | what's left when the remainder vanishes |
The remainder term is also the error — it tells you the curve sits 3/(x+1) above the line, so it approaches from above as x→+∞ and from below as x→−∞.
√(x²) = |x|, which is −x when x is negative. Pulling x out of a root as x→−∞ introduces a minus sign.
| limx→−∞ √(4x²+1)/x | |
| = lim |x|·√(4 + 1/x²) / x | factor x² out of the root — it exits as |x|, not x |
| = lim (−x)·√(4 + 1/x²) / x | x is negative, so |x| = −x |
| = −2 | the x's cancel, leaving −√4 |
The same limit as x→+∞ is +2. This sign is the most-missed item in the unit.
The growth hierarchy
"≪" means the ratio goes to 0 — the right-hand one utterly swamps the left. Any ratio of two of these has an answer you can read off without work.
ln x ≪ xp: logarithms grow like the number of digits. Going from a thousand to a trillion multiplies x by a billion and raises ln x by about 21. Nothing polynomial can be beaten that slowly.
xp ≪ ex: differentiating a polynomial lowers the degree; do it p+1 times and it's gone. Differentiating ex changes nothing. Apply L'Hôpital enough times and the polynomial dies while the exponential stands there.
ex ≪ x!: ex multiplies by a fixed e at each step; x! multiplies by an ever-growing factor. Fixed ratio always loses to growing ratio.
x! ≪ xx: x! is x·(x−1)·(x−2)···, all factors below x; xx is x multiplied by itself x times.
Worked — reading answers straight off the hierarchy
| limx→∞ (ln x)/x | ln loses to any power → 0 |
| limx→∞ x100/ex | any polynomial loses to ex, no matter how big the exponent → 0 |
| limx→∞ ex/x100 | same fact upside down → ∞ |
| limx→∞ (x³ + ln x)/(2x³ − ex) | on top x³ dominates ln x; on the bottom ex dominates x³. So this behaves like x³/(−ex) → 0 |
The technique for a messy expression: in each of the numerator and denominator separately, keep only the fastest-growing term and throw everything else away. Then compare the two survivors.
Show limx→∞ x²/ex = 0 rather than taking it on faith.
| x²/ex is ∞/∞ | both blow up; indeterminate, so L'Hôpital applies |
| → 2x/ex | differentiate top and bottom separately; the polynomial dropped a degree, the exponential didn't budge |
| still ∞/∞ → 2/ex | again; now the top is a constant |
| = 0 | constant over something blowing up |
Notice the mechanism: each round of differentiation costs the polynomial one degree and costs ex nothing. A degree-100 polynomial just takes 100 rounds. It always loses eventually — which is what "≪" is asserting.
It's not a limit trick; it's the ranking of how fast things can grow, and it recurs constantly:
- In this course, Unit 10: whether an infinite series converges is entirely a question of whether its terms shrink fast enough. Every convergence test is the hierarchy in disguise.
- Improper integrals: ∫₁∞ dx/xp converges only when p > 1. The threshold is a growth-rate threshold.
- Computer science: an algorithm that takes ln n steps is fine at any scale; n² is usable; 2n is unusable past about n = 50; n! is unusable past about n = 15. Same ordering, and it's the reason some problems are considered intractable.
- Compound growth generally: exponential eventually beats polynomial always, however unfavourable the constants. A quantity growing 1% a year overtakes any fixed-power trend, given enough time. That's the mathematical content of most arguments about compounding.
Also useful as a sanity check: if an answer implies a polynomial outran an exponential, the arithmetic is wrong.
∞ − ∞ with roots: rationalize
| limx→∞ (√(x²+x) − x) | ∞−∞, indeterminate — both pieces blow up |
| = lim (x²+x−x²)/(√(x²+x)+x) | multiply by the conjugate over itself |
| = lim x/(√(x²+x) + x) | numerator collapses |
| = lim 1/(√(1 + 1/x) + 1) | divide top and bottom by x (positive here, so |x| = x) |
| = 1/2 | the 1/x → 0 |
L'Hôpital's rule
Most BC courses don't teach this until Unit 4 (applications of the derivative), because the rule uses derivatives and she won't have them yet in September. I've put it here because it belongs with the limit toolkit conceptually, and because you'll want it when she brings home a hard limit.
Don't front-run her teacher with it. If she's in the limits unit and reaches for L'Hôpital on a problem meant to be done by factoring, she'll lose points for using a tool that isn't in scope yet — and more importantly she'll skip the algebra practice the unit exists to build.
It comes back properly in Unit 4 with a full set of worked examples and the error cases. Read this now for your own map; use it with her later.
If the limit has form 0/0 or ∞/∞, then lim f/g = lim f′/g′.
Near the point a, each function is well approximated by its tangent line. If f(a) = g(a) = 0, those tangent lines both pass through zero there, so near a:
The ratio is then f′(a)(x−a) / g′(a)(x−a), and the (x−a) cancels. What's left is f′(a)/g′(a).
L'Hôpital is just cancelling the common factor again — the same move as every algebraic technique above, done with tangent lines instead of factoring. That's why it needs the 0/0 condition: without it, the tangent lines don't both pass through zero and the cancellation isn't available.
Historical aside worth having: L'Hôpital didn't discover it. He paid Johann Bernoulli a retainer for exclusive rights to his mathematical output, published it in the first-ever calculus textbook in 1696, and credited him only vaguely. Bernoulli complained about it for the rest of his life.
One: this is not the quotient rule. Differentiate numerator and denominator separately. You are not differentiating the fraction.
Two: verify the form is 0/0 or ∞/∞ before applying. Used on something like 2/0, it produces a confidently wrong answer with no warning.
Rewriting the other indeterminate forms
| Form | Move |
|---|---|
| 0 · ∞ | Send one factor to the denominator: x·ln x = (ln x)/(1/x), now ∞/∞ |
| ∞ − ∞ | Common denominator, or rationalize |
| 1∞, 00, ∞0 | Take ln, find the limit of the log, then exponentiate |
- ln(ab) = ln a + ln b
- ln(a/b) = ln a − ln b
- ln(ab) = b · ln a — the one that does the work, because it drags an exponent down to ground level where you can differentiate it
And the inverse pair: eln x = x and ln(ex) = x. Also ax = ex ln a, which is how any exponential gets differentiated — convert to base e and chain rule.
Example: ln(x⁵) = 5 ln x. An exponent became a coefficient.
Find limx→0+ x · ln x. The first factor goes to 0, the second to −∞. Neither wins by inspection.
| x · ln x | form 0·(−∞); L'Hôpital needs a fraction, so build one |
| = (ln x)/(1/x) | move x downstairs as its reciprocal — now −∞/∞ |
| → (1/x) / (−1/x²) | L'Hôpital: derivative of ln x is 1/x; derivative of x−1 is −x−2 |
| = (1/x)·(−x²/1) = −x | dividing by a fraction = multiplying by its reciprocal |
| = 0 |
You had a choice of which factor to send downstairs. Sending ln x down instead would give x/(1/ln x), which L'Hôpital turns into something worse. Move the factor whose reciprocal is simpler — usually the algebraic one, not the log.
Find limx→0+ ( 1/x − 1/sin x ). Both terms blow up to +∞; the question is whether the difference settles.
The whole thing comes out of the same inequality chain that produced sin x / x = 1 — go back to the sector picture, where we had sin x ≤ x ≤ tan x. Two facts fall out of it:
| 1/x − 1/sin x = (sin x − x)/(x sin x) | common denominator — always the first move for ∞−∞ |
| sin x − x ≤ 0, and x sin x > 0 | first inequality; so the whole expression is ≤ 0. That's the upper bound. |
| sin x − x ≥ x cos x − x = x(cos x − 1) | second inequality, minus x on both sides |
| so (sin x − x)/(x sin x) ≥ x(cos x − 1)/(x sin x) | replacing the numerator by something smaller can only lower the fraction |
| = (cos x − 1)/sin x | the x cancels. That's the lower bound. |
| = −[(1 − cos x)/x] · [x/sin x] | multiply and divide by x to manufacture two known limits |
| → −(0)·(1) = 0 | the special limits: (1−cos x)/x → 0 and x/sin x → 1 |
| squeezed between 0 and something → 0, so the limit is 0 | Squeeze Theorem |
Two infinities cancelling exactly. Sanity-check it numerically: at x = 0.1 the expression is about −0.0167, at x = 0.01 about −0.00167 — shrinking by a factor of ten each time, so it's heading to zero linearly.
This is worth doing the long way because it shows the sector diagram is not a one-trick tool. That single inequality chain — sin x ≤ x ≤ tan x — is the source of essentially every trig limit in the unit. If you remember the picture, you can regenerate the bounds; if you have the bounds, you can squeeze almost anything.
Find limx→∞ (1 + 3/x)x. The base → 1 and the exponent → ∞. Tempting to say 1 (since 1 to any power is 1) or ∞ (since it's more than 1, compounding forever). Both are wrong; this is genuinely indeterminate.
| y = (1 + 3/x)x | name it so you can take logs |
| ln y = x · ln(1 + 3/x) | log law brings the exponent down; now ∞·0 |
| = ln(1 + 3/x) / (1/x) | make it a fraction — now 0/0 |
| → [ (1/(1+3/x)) · (−3/x²) ] / (−1/x²) | L'Hôpital; the top needs the chain rule |
| = 3/(1 + 3/x) | the −1/x² cancels top and bottom |
| → 3 | as x→∞, 3/x → 0 |
| y → e³ | exponentiate to undo the log |
This is the compound-interest limit from earlier, arrived at by machinery instead of by definition. Consistent, which is reassuring: lim(1 + k/x)x = ek.
Find limx→∞ x1/x. Base → ∞, exponent → 0.
| y = x1/x | |
| ln y = (1/x)·ln x = (ln x)/x | exponent down; now ∞/∞ |
| → (1/x)/1 = 1/x | L'Hôpital |
| → 0 | |
| y → e0 = 1 | exponentiate |
Sanity check against the growth hierarchy: ln x ≪ x, so (ln x)/x → 0, so the whole thing → e⁰ = 1. Same answer, no calculus. The hierarchy is faster when it applies.
Find limx→∞ x/√(x²+1).
| form is ∞/∞, so L'Hôpital is legal | |
| → 1 / [x/√(x²+1)] | differentiate top and bottom |
| = √(x²+1)/x | which is the reciprocal of what we started with |
| applying it again returns the original | infinite loop — L'Hôpital never terminates here |
| Do it by algebra: divide by x → 1/√(1 + 1/x²) → 1 |
L'Hôpital being legal doesn't make it the right tool. If two applications haven't simplified anything, stop and look for algebra.
| Find limx→0+ xx. Set y = xx. | variable in both base and exponent → take logs |
| ln y = x · ln x | the exponent comes down; form is now 0·(−∞) |
| = (ln x)/(1/x) | move x to the denominator as 1/x — now ∞/∞ |
| → (1/x)/(−1/x²) | L'Hôpital: differentiate top and bottom separately |
| = −x → 0 | simplify the compound fraction |
| y → e0 = 1 | we found the limit of ln y; exponentiate to recover y |
Finding ln y → 0 and answering "0". You found the limit of the logarithm. The last step is never optional.
Continuity
f is continuous at a when limx→a f(x) = f(a). Three claims bundled into one equation: the limit exists, f(a) exists, and they agree.
Continuity means taking the limit and evaluating the function commute — you can do them in either order and get the same thing. That's the useful form, because it's exactly what licenses "just substitute" as a limit technique, and it's what lets you pass a limit inside a continuous function later on.
"Find k making f continuous"
Set left-hand limit = right-hand limit = f(a), and solve. If there are two unknowns, the second equation almost always comes from also requiring differentiability — match the derivatives of the two pieces as well as their values.
The Intermediate Value Theorem
If f is continuous on [a, b] and N is any value between f(a) and f(b), then f(c) = N for some c in [a, b].
Informally: a continuous curve can't get from below a line to above it without crossing it. Obvious — for the reals.
It is false over the rationals. Take f(x) = x² − 2 on [1, 2] with only rational inputs allowed. f(1) = −1, f(2) = 2, so it must cross zero — but the crossing point is √2, which isn't rational. The function jumps over zero without ever landing on it.
So IVT isn't a fact about continuity alone. It's a fact about continuity plus the completeness of the reals. This is the payoff of that earlier note about gaps, and it's the reason existence theorems in calculus are theorems rather than observations.
Show that x³ − 4x + 1 = 0 has a solution between 0 and 1.
| f(x) = x³ − 4x + 1 is a polynomial, so continuous on [0,1] | state this explicitly — it's a scored step |
| f(0) = 0 − 0 + 1 = 1 | positive |
| f(1) = 1 − 4 + 1 = −2 | negative |
| 0 lies between −2 and 1 | the value we want is bracketed |
| By the IVT, there is a c in (0,1) with f(c) = 0 | name the theorem in the conclusion |
IVT proves a root exists. It gives you no way to find it, and it doesn't say the root is unique — there might be several. "At least one" is all you can ever claim.
On AP free response, you must explicitly state that f is continuous and name the Intermediate Value Theorem. Doing the arithmetic without the justification sentence scores zero on that part. Same pattern applies to MVT and EVT later.
Weierstrass's function (1872): continuous everywhere, differentiable nowhere. A curve with no smooth point anywhere — no tangent line at any location, infinitely wrinkled at every scale.
Before it, mathematicians assumed continuity basically implied smoothness apart from isolated corners. Hermite called it "a lamentable plague." It's the ancestor of fractals and of Brownian motion — actual stock price paths and actual pollen-grain trajectories are exactly this kind of object. The intuition "continuous means you can draw it without lifting the pen, so it must have a direction almost everywhere" is false, and this is the counterexample.
Next stop — real analysis. Redo the entire year with proofs. Where ε-δ is the point, where completeness gets stated honestly, and where the pathological examples (Weierstrass's function, the Cantor set) stop being curiosities and become the objects that force the definitions.
Then — measure theory and the Lebesgue integral, which fixes the Riemann integral's limitations and is the actual foundation of modern probability. Also topology, which is what you get when you keep "nearness" and throw away distance and number entirely.
And the wildcard: nonstandard analysis (Robinson, 1966), which vindicates Leibniz by constructing infinitesimals rigorously. If she ever objects that treating dy/dx as a fraction "shouldn't be allowed," this is the field that says she was right the whole time.
An actual ε-δ proof. BC states the definition and never uses it. A university Calc I makes you prove a few, and doing one is worth an hour because it shows the definition is a procedure, not a decoration.
Prove that limx→2(3x + 1) = 7.
| Let ε > 0 be given. | the challenger moves first; ε is arbitrary |
| We need |(3x+1) − 7| < ε. | write down the goal |
| |3x − 6| = 3|x − 2| | work backwards from the goal to the hypothesis — this is the whole technique |
| 3|x−2| < ε ⟺ |x−2| < ε/3 | so the answer is staring at us |
| Choose δ = ε/3. Then 0 < |x−2| < δ ⟹ |3x+1−7| = 3|x−2| < 3δ = ε. ∎ |
The pattern generalizes: manipulate the thing you want small until the factor |x − a| appears, then read off δ. For nonlinear functions you also have to bound the other factor first — for f(x) = x², you'd first restrict to |x−2| < 1 so that |x+2| < 5, then take δ = min(1, ε/5). That "min" is the signature of every ε-δ proof you'll ever see.
The Russian tradition treats limits as algebra problems and gets a great deal further without any calculus machinery. The habit is worth stealing: L'Hôpital is often the slowest route, and it isn't available during the limits unit anyway.
Find limx→0 (∛(1+x) − 1)/x, using nothing but algebra.
| set a = ∛(1+x), b = 1, so a³ = 1+x and b³ = 1 | name the pieces |
| recall a³ − b³ = (a − b)(a² + ab + b²) | the cube analogue of the conjugate trick |
| multiply top and bottom by (a² + a + 1) | this is the "cubic conjugate" |
| numerator becomes a³ − 1 = (1+x) − 1 = x | the root is gone |
| = x / [x(a² + a + 1)] = 1/(a² + a + 1) | cancel |
| → 1/(1 + 1 + 1) = 1/3 | as x→0, a→1 |
Why this is the better habit: the same move handles any root — for fifth roots, multiply by a⁴+a³+a²+a+1. And notice the answer: 1/3 is exactly the derivative of x1/3 at x = 1, which she'll confirm in Unit 2. The algebra found the derivative before the derivative existed.
Try next, same spirit: limx→0 (√(1+x) − √(1−x))/x. Answer 1, by ordinary conjugate.
Unit 1 — Limits and Continuity
Definition
- limx→a f(x) = L: for every ε > 0 there is δ > 0 with |f(x) − L| < ε whenever 0 < |x − a| < δ
- Two-sided limit exists ⟺ both one-sided limits exist and agree
- Continuous at a ⟺ limx→a f(x) = f(a) (limit exists, f(a) exists, they match)
The five special limits
- limx→0 (sin x)/x = 1 — sector squeeze; radians only
- limx→0 (1 − cos x)/x = 0 — conjugate
- limx→0 (1 − cos x)/x² = 1/2 — conjugate, split the x²
- limx→0 (tan x)/x = 1 — = (sin x/x)(1/cos x)
- limx→0 (ex − 1)/x = 1 — definition of e
- limn→∞ (1 + x/n)n = ex — continuous compounding
0/0 toolkit
- Factor and cancel — the hidden (x−a) is guaranteed by the Factor Theorem
- Rationalize with the conjugate — whenever a root appears
- Combine complex fractions — clean the numerator into one fraction first
- Recognize a difference quotient — [f(a+h) − f(a)]/h is just f′(a)
Limits at infinity (rational functions)
- Top degree > bottom → ±∞ (slant asymptote if it exceeds by exactly 1)
- Top degree < bottom → 0
- Equal degrees → ratio of leading coefficients
- Method: divide every term by the highest power in the denominator
- √(x²) = |x| — this is −x when x < 0
Growth hierarchy
- ln x ≪ xp ≪ ex ≪ x! ≪ xx
- Keep only the fastest-growing term in numerator and denominator, then compare
Indeterminate forms and the move
- 0/0, ∞/∞ — algebra first; L'Hôpital if algebra fails
- 0·∞ — send one factor downstairs as a reciprocal
- ∞−∞ — common denominator, or rationalize
- 1∞, 00, ∞0 — take ln, find the limit, then exponentiate
- Not indeterminate: nonzero/0 (blows up), 0/nonzero (equals 0)
Theorems, with the phrases that score
- Squeeze: g ≤ f ≤ h and g, h → L, then f → L
- IVT: f continuous on [a,b], N between f(a) and f(b) ⟹ f(c) = N for some c. Say "continuous" and say "Intermediate Value Theorem."
Four discontinuities
- Removable (hole) · Jump (sides disagree) · Infinite (asymptote) · Oscillating
- Only removable ones can be repaired by redefining a single point
Practise Unit 1 → Fresh problems from this unit, with worked solutions.