Before we can differentiate or integrate anything, we need a precise way to talk about what a function approaches. A limit answers the question: as the input x slides toward some value a, does the output f(x) settle near a single number L? Crucially, the limit ignores what happens atx=a itself; it cares only about the neighborhood around .
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Lesson close · Calculus for Quants
Consolidate Limits & Continuity
Use this checkpoint to recall the lesson before marking it complete. The concepts below come from this lesson's authored sections.
Key concepts covered
01The idea of a limit
02Formal definition (epsilon-delta)
03A worked epsilon-delta derivation
04One-sided limits and limits at infinity
05Continuity and its failures
What you should know now
□Explain Limits & Continuity in your own words and state the assumptions.
□Reproduce the main argument, derivation, or implementation without copying the lesson.
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Check one example against units, boundary conditions, or edge cases.
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Sequence
Continue to The Derivative & Differentiation Rules, the next prerequisite step in this course.
This distinction matters for quants. A bond price as time-to-maturity shrinks, an option's implied volatility as the strike approaches the forward, a Monte Carlo estimator as the path count grows: each is a limit, and each may behave badly exactly at the point of interest while being perfectly well-behaved nearby.
Formal definition (epsilon-delta)
The rigorous definition pins down "approaches" with two tolerances.
limx→af(x)=L⟺∀ε>0∃δ>0s.t.0<∣x−a∣<δ⟹∣f(x)−L∣<ε.
Read it as a game: you name an output tolerance ε, and I must produce an input tolerance δ so that every x within δ of a (but not equal to a) lands within ε of L. If I can always win, the limit is L.
A worked epsilon-delta derivation
Claim: limx→3(2x+1)=7. Given ε>0, we need δ so that ∣x−3∣<δ forces ∣(2x+1)−7∣<ε. Now
∣(2x+1)−7∣=∣2x−6∣=2∣x−3∣.
We want 2∣x−3∣<ε, i.e. ∣x−3∣<ε/2. So choose δ=ε/2. Then ∣x−3∣<δ gives ∣(2x+1)−7∣=2∣x−3∣<2δ=ε. The game is won for every ε, so the limit is exactly 7.
One-sided limits and limits at infinity
Sometimes the approach direction matters. The right-hand limitlimx→a+f(x) restricts to x>a; the left-hand limitlimx→a−f(x) restricts to x<a. The two-sided limit exists if and only if both one-sided limits exist and agree.
Example: for f(x)=x/∣x∣, the right limit at 0 is +1 and the left limit is −1, so limx→0f(x) does not exist.
A limit at infinity describes long-run behavior: limx→∞f(x)=L means f(x) can be forced within ε of L for all sufficiently large x. Formally, for every ε>0 there is an M such that x>M implies ∣f(x)−L∣<ε; the threshold M plays the role δ played before. For instance
limx→∞x2−13x2+5=3,
found by dividing numerator and denominator by x2 and noting the 5/x2 and 1/x2 terms vanish. The same technique handles the asymptotic yield of a perpetuity or the limiting price of a long-dated cash flow.
Limits also obey clean algebraic laws that let us avoid epsilon-delta for routine work. If limx→af(x) and limx→ag(x) both exist, then the limit of a sum, difference, product, or quotient (denominator limit nonzero) is the corresponding combination of the limits. These laws, provable from the definition, are why we can compute most limits by substitution once continuity is established.
Continuity and its failures
A function f is continuous at a when the limit exists, the value exists, and they coincide:
limx→af(x)=f(a).
Three standard ways this breaks:
Removable: the limit exists but differs from (or is missing) the value, e.g. g(x)=x−1x2−1 at x=1, where the limit is 2 but g(1) is undefined. Redefining the point patches the hole.
Jump: left and right limits exist but disagree, as with x/∣x∣ at 0.
Infinite (essential): a limit blows up, e.g. 1/x at 0.
Intermediate Value Theorem
If f is continuous on a closed interval [a,b] and N lies between f(a) and f(b), then there exists c in (a,b) with f(c)=N. Concretely, take f(x)=x3−x−1. Since f(1)=−1<0 and f(2)=5>0, continuity guarantees a root somewhere in (1,2). This is the theoretical engine behind bisection root-finding.
Why quants care
Continuity is the license to use calculus at all. Pricing functions must be continuous in their inputs for hedge ratios (derivatives) to exist and for interpolation of yield curves to be sound. The IVT underpins bisection, the workhorse for inverting Black-Scholes to recover implied volatility from a market price: because price is continuous and monotone in volatility, a bracketed root must exist and can be squeezed numerically. Jump discontinuities flag genuine modeling hazards: a payoff with a jump (a digital option) cannot be delta-hedged smoothly near the strike, which is exactly where risk concentrates.
Adapted from MIT OpenCourseWare (18.01 / 18.02 Calculus), CC BY-NC-SA.
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Knowledge check
Q1.In the epsilon-delta definition, what is the logical order of quantifiers?
Q2.Why does the two-sided limit of x/|x| fail to exist at 0?
Q3.The function (x^2-1)/(x-1) at x=1 has which kind of discontinuity?
Q4.The Intermediate Value Theorem guarantees a root of x^3 - x - 1 in (1,2) because:
Problems
P1
Compute the limit as x approaches infinity of (3x^2 + 5)/(x^2 - 1). Give a decimal.
P2
Using the epsilon-delta definition for lim_{x->3}(2x+1)=7, find the largest delta that works for epsilon = 0.1. Give a decimal.
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