Interactive Math & Science Lessons
LIMIT THEORY — WHAT THE GRAPH DOES AS x CLOSES IN ON a
A limit asks where f is heading as x approaches a, never what f does at a. Read the two one-sided limits first: if they agree you have a limit, and the only question left is whether f(a) matches it. If they disagree, the way they disagree — a step, a blow-up, or an endless wobble — names the case.
01
What you know: The left and right limits are equal and finite, but f(a) is undefined. Substituting a gives 0/0, and no factor cancels.
Test: L⁻ = L⁺ ≠ f(a)
Worked example: f(x) = sin(x)/x at a = 0 → f(0) reads 0/0, yet f(±0.001) = 0.99999983 → limit = 1
Open this case: sinc sin(x)/x
02
What you know: Substituting a gives 0/0, and the numerator and denominator share a factor of (x − a).
Test: 0/0 → ÷(x − a)
Worked example: f(x) = (x² − 1)/(x − 1) at a = 1 → (x − 1)(x + 1)/(x − 1) = x + 1 → limit = 2
Open this case: L'Hopital
03
What you know: Both one-sided limits are finite, but they are different numbers. The graph steps from one level to another at a.
Test: L⁻ ≠ L⁺
Worked example: f(x) = |x|/x at a = 0 → left limit = −1, right limit = +1 → the limit does not exist
Open this case: Jump |x|/x
04
What you know: At least one side grows without bound. Substituting a gives a non-zero number over zero, so a vertical asymptote stands at x = a.
Test: f → ±∞ at x = a
Worked example: f(x) = 1/x at a = 0 → values → −∞ from the left and → +∞ from the right → no finite limit
Open this case: Infinite 1/x
05
What you know: The function stays bounded, yet keeps sweeping the same range infinitely often, however close to a you look.
Test: ∄L, |f| ≤ 1
Worked example: f(x) = sin(1/x) at a = 0 → between x = 0.001 and x = 0.002 the graph completes about 80 full cycles, hitting every value in [−1, 1]
Open this case: Oscillation sin(1/x)
Prove that sin x / x → 1 by producing a δ for every ε, and predict the table's two usable rows before reading them. This is f(x) = sin(x)/x with x → 0.
Dividing the sine series by x is legal for every x ≠ 0, which is the only place a limit ever looks. What comes out has nothing but even powers of x in it, and that is why the left and right columns of the table carry identical digits on every row.
Two terms are enough here: the x⁴ term is already down at 8.33 × 10⁻⁷, too small to touch the sixth decimal.
Move in by a factor of ten and the gap to 1 shrinks by a factor of a hundred. The approach is quadratic in x, which is a much stronger claim than “the limit is 1” — and the table's own two rows are enough to establish it.
Evidence is not proof, so bound the whole tail at once. For |x| ≤ 1 the series alternates with shrinking terms, so everything after the 1 is smaller in size than the first term dropped, and that term is x²/6. Turning the inequality around makes δ a formula rather than a search.
Read what the square root costs. Tightening ε by a factor of a million buys back only a factor of a thousand in x.
Answer
The table prints 0.998334 at x = 0.1 and 0.999983 at x = 0.01, and both were written out here from four terms of a series before the panel was consulted. What δ = √(6ε) adds is certainty: name any tolerance and the x that meets it is one square root away, so no amount of further zooming can turn up a surprise. It also explains why the table runs out of things to say. Its rows step x down by ten while the gap falls by a hundred, so by the third row that gap is 1.6667 × 10⁻⁷ — under half a unit in the sixth decimal — and that row and the five below it all print 1.000000, though the value at x = 0.001 is 0.99999983. Keep the gap rather than discarding it: divide the first row's 1.665834 × 10⁻³ by 0.01 and you get 0.16658, which is 1/6. A second limit sits underneath the famous one, and the series named it in advance.
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