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Mathematics A — Week 3
HSG · Autumn Semester 2026 · Monastery Masterclass
Limits & Continuity: When Functions Behave
Guided learning · 10–15 minutes · Before Exercise Series 2
Imagine you're analyzing a company's revenue as its price changes. A smooth change in price should normally produce a predictable change in revenue. But what if your model suddenly jumps, becomes undefined, or gives completely different predictions depending on which direction you approach a price?
This week, you'll learn how mathematics handles these situations.
The central idea is simple:
A limit tells us where a function is heading. Continuity tells us whether the function actually reaches that value without a break.
By the end of this masterclass, you should be able to determine domains and ranges, recognize when inverse functions exist, work with exponential and logarithmic functions, calculate limits, classify discontinuities, and use Bolzano's theorem to establish whether an equation has a solution.
This lesson builds on the official Lecture 3 topic, Limits of Functions and Continuity, including the function foundations needed for Exercise Series 2.
1. Functions: Know what you're working with
The foundation · 2 minutes
A function is a machine: you provide an input \(x\), and it returns exactly one output \(f(x)\).
But three things matter:
- Domain \(D_f\): which inputs are allowed?
- Codomain: the set in which outputs are declared to lie.
- Range \(W_f\): which outputs are actually produced?
For example, consider:
\[ f(x)=\sqrt{x-2} \]
We cannot take the real square root of a negative number, so \(x\geq2\). And a square root never produces a negative output.
\[ \boxed{D_f=[2,\infty),\qquad W_f=[0,\infty)} \]
Notice the difference: the domain concerns inputs; the range concerns outputs.
Three domain restrictions to recognize
A composite expression must satisfy all relevant restrictions simultaneously.
Inverse functions: Running the machine backwards
Suppose you know the output of a function and want to recover its input.
That is what the inverse function \(f^{-1}\) does.
For example:
\[ f(x)=2x+3 \]
Reverse the operations: subtract 3, then divide by 2.
\[ \boxed{f^{-1}(x)=\frac{x-3}{2}} \]
The inverse swaps inputs and outputs, so:
\[ D_{f^{-1}}=W_f,\qquad W_{f^{-1}}=D_f \]
The important condition: different inputs must not produce the same output. Otherwise, the inverse would not know which input to return. This property is called injectivity.
Consider \(f(x)=x^2\) on \(\mathbb R\). Both \(2\) and \(-2\) give \(4\), so there is no inverse function on the whole real-number domain. Restrict the domain to \([0,\infty)\), however, and the inverse becomes \(f^{-1}(x)=\sqrt{x}\).
Exam trap: \(f^{-1}(x)\) means the inverse function, not \(1/f(x)\). Also, strictly increasing or strictly decreasing functions are injective, but a function need not be strictly monotonic to be injective.
Three other function properties are useful to recognize before studying continuity:
These properties become especially useful for proving that solutions are unique and, in later weeks, for optimization.
2. Exponentials and logarithms: Opposite operations
Essential function families · 2 minutes
Exponential functions describe growth or decay. Logarithms reverse that process.
Imagine a market growing exponentially:
\[ V(t)=100e^{0.2t} \]
Here, \(V(0)=100\). As time passes, the value increases.
But suppose we want to know when the value reaches 150.
\[ \begin{aligned} 150&=100e^{0.2t}\\ 1.5&=e^{0.2t}\\ \ln(1.5)&=0.2t \end{aligned} \]
Therefore:
\[ \boxed{t=\frac{\ln(1.5)}{0.2}\approx2.027} \]
We used the natural logarithm \(\ln\) to reverse the exponential.
The essential relationship
For \(a>0\), \(a\neq1\):
\[ \boxed{y=a^x\iff x=\log_a(y),\quad y>0} \]
Exponential: \(a^x\)
Input any real number; output always positive
Illustration for \(a>1\)
Logarithm: \(\log_a(x)\)
Input positive numbers only; output any real number
Illustration for \(a>1\)
For \(a>1\), both functions are strictly increasing. For \(0\<a<1\), both are strictly decreasing. Either way, each has an inverse.
Two relationships you should understand:
\[ \ln(e^x)=x,\qquad e^{\ln x}=x\quad(x>0) \]
And the fundamental logarithm rules, for \(a,b>0\):
\[ \ln(ab)=\ln a+\ln b \]
\[ \ln\left(\frac ab\right)=\ln a-\ln b \]
\[ \ln(a^r)=r\ln a \]
These identities are not printed on the supplied Mathematics A exam formulary. Learn them or be able to derive them using exponential laws.
Exam trap: \(\ln(a+b)\neq\ln a+\ln b\) in general. Logarithms turn multiplication into addition, not addition into addition.
3. Limits: Where is the function heading?
The central concept · 2 minutes
Imagine moving towards a point on a graph.
A limit is about what happens as you approach the point, not necessarily what happens exactly at the point.
We write:
\[ \lim_{x\to x_0}f(x)=L \]
This means that as \(x\) gets arbitrarily close to \(x_0\) (while remaining different from it), \(f(x)\) gets arbitrarily close to \(L\).
Approaching from two directions
Approaching \(x_0=2\)
From the left
1.9 → 1.99 →
\(\lim_{x\to2^-}f(x)\)
2
Target point
From the right
← 2.01 ← 2.1
\(\lim_{x\to2^+}f(x)\)
A finite two-sided limit exists exactly when both one-sided limits exist and are equal.
Mathematically:
\[ \boxed{ \lim_{x\to x_0^-}f(x) = \lim_{x\to x_0^+}f(x) =L } \]
If the two sides approach different values, the two-sided limit does not exist.
Worked example: A hole in a function
Consider:
\[ f(x)=\frac{x^2-4}{x-2},\qquad x\neq2 \]
What happens as \(x\to2\)?
Direct substitution gives \(0/0\), which is indeterminate. We must simplify.
Factor the numerator:
\[ \begin{aligned} f(x)&=\frac{(x-2)(x+2)}{x-2}\\ &=x+2,\qquad x\neq2 \end{aligned} \]
So:
\[ \boxed{\lim_{x\to2}f(x)=4} \]
The value \(f(2)\) is undefined, but the limit exists.
This illustrates an important distinction: the function's behavior near a point can be perfectly predictable even when its value at the point is missing.
4. Continuity: Does the function meet its limit?
The main exam concept · 2 minutes
Think of continuity as a promise:
As you approach a point, the function approaches the value it actually takes there.
For an interior point \(x_0\) of the domain, continuity requires three conditions.
1. \(f(x_0)\) is defined.
2. The left and right finite limits exist and agree.
3. Their common limit equals \(f(x_0)\).
\[ \boxed{\lim_{x\to x_0}f(x)=f(x_0)} \]
At endpoints of the domain, continuity is checked using the relevant one-sided limit within the domain.
Four ways continuity can fail
Technical distinction: Continuity is formally defined at points in a function's domain. An excluded point may be called a domain gap or discontinuity in broader terminology; a function can still be continuous everywhere on its actual domain.
Only the removable type can be repaired simply by defining or changing the value at that one point.
Worked example: Can two formulas connect smoothly?
Suppose a company uses the following revenue model:
\[ R(x)= \begin{cases} 2x+1,&x<1\\ x+2,&x\geq1 \end{cases} \]
Is revenue continuous at \(x=1\)?
From the left:
\[ \lim_{x\to1^-}R(x)=2(1)+1=3 \]
From the right:
\[ \lim_{x\to1^+}R(x)=1+2=3 \]
And the actual value is \(R(1)=3\).
All three agree, so the function is continuous at \(x=1\).
Even though it uses different formulas on each side, there is no jump.
Continuity of combined functions
Polynomials, exponentials, logarithms and trigonometric functions are continuous on their respective natural domains. Sums, products, quotients with nonzero denominators, and compositions of continuous functions remain continuous wherever the operations are defined.
For example:
\[ g(x)=\ln(x^2+1) \]
Since \(x^2+1>0\) for every real \(x\), the logarithm is always defined and the composition is continuous everywhere.
Therefore:
\[ \lim_{x\to2}\ln(x^2+1)=\ln5 \]
No special limit technique is needed.
5. Calculating limits: Choose the right method
The problem-solving toolkit · 2 minutes
Most introductory limit questions can be approached with a short decision process.
Start: Can you substitute \(x_0\) directly?
Yes — defined, continuous
Evaluate \(f(x_0)\)
No — undefined or indeterminate
Simplify, rationalize, or use a special limit
Here are the most useful techniques:
Example: Using the exam formulary
The official Mathematics A formulary directly provides:
\[ \lim_{u\to0}\frac{e^u-1}{u}=1 \]
Now evaluate:
\[ \lim_{x\to0}\frac{e^{3x}-1}{x} \]
Let \(u=3x\), so \(u\to0\) as \(x\to0\).
\[ \begin{aligned} \lim_{x\to0}\frac{e^{3x}-1}{x} &=3\lim_{x\to0}\frac{e^{3x}-1}{3x}\\ &=3\cdot1\\ &=\boxed{3} \end{aligned} \]
The original special limit is printed; the transformed version is derived, not separately printed.
Two other growth comparisons printed on the formulary are:
\[ \lim_{x\to\infty}\frac{x^k}{e^x}=0\quad(k\in\mathbb Z) \]
\[ \lim_{x\to\infty}\frac{\ln x}{x^r}=0\quad(r>0) \]
Their intuition is worth remembering:
Exponential growth eventually dominates every positive integer power, and positive powers eventually dominate logarithmic growth.
6. Bolzano's theorem: Proving a solution exists
A powerful economic application · 2 minutes
Suppose you know that a continuous function is negative at one price and positive at another.
Does it have to cross zero somewhere in between?
Yes.
Bolzano's theorem states that if \(f\) is continuous on \([a,b]\) and
\[ f(a)\cdot f(b)<0, \]
then there is at least one \(c\in(a,b)\) such that:
\[ \boxed{f(c)=0} \]
Visual intuition: A continuous curve must cross zero to change sign.
Original schematic of the intermediate value principle; not to scale.
Worked example: Existence versus uniqueness
Consider:
\[ f(x)=x^3+x-1 \]
We want to know whether \(f(x)=0\) has a solution between 0 and 1.
First check the signs:
\[ f(0)=-1<0,\qquad f(1)=1>0 \]
As a polynomial, \(f\) is continuous everywhere.
Therefore Bolzano guarantees at least one root inside \((0,1)\).
But what about uniqueness?
Both \(x^3\) and \(x\) are strictly increasing. Their sum is strictly increasing too. Subtracting 1 does not change that.
A strictly increasing function can cross zero at most once.
Therefore, there is exactly one root in \((0,1)\).
This distinction is crucial:
\[ \underbrace{\text{Continuity + sign change}}_{\text{Existence}} \qquad+\qquad \underbrace{\text{Strict monotonicity}}_{\text{Uniqueness}} \]
Bolzano proves existence. Strict monotonicity is an additional way to prove uniqueness.
It also connects directly to business and economics: when excess demand changes sign between two prices, continuity can establish the existence of a market-clearing price.
7. Your exam toolbox
The exam provides the official Mathematics A formulary. That does not mean every important formula or theorem is printed.
The official formulary does not provide a general ready-to-use formula for evaluating arbitrary limits.
Five mistakes that cost easy points
- Treating \(0/0\) as a numerical answer rather than an indeterminate form.
- Calculating only one side of a limit when the function is piecewise.
- Assuming a function is continuous because its two-sided limit exists, without checking its actual value.
- Ignoring domain restrictions after simplifying a fraction or taking a logarithm.
- Claiming that Bolzano guarantees exactly one solution without establishing uniqueness.
Knowledge check
Apply the ideas before unlocking the exercises. Allow approximately 4–6 minutes, then compare with the explained answer key.
1
Domain and continuity
Consider f(x) = ln(4 − x²). Determine its domain. Is it continuous on that domain? Could you define f(2) as a finite number to make it continuous at x = 2?
2
Can a jump be repaired?
A function equals x + 1 for x < 0 and 3 − x for x ≥ 0. Can you make it continuous at zero by changing only f(0)? Explain.
3
Inverse functions with a restricted domain
Let f(x) = x² − 4x with domain [2,∞). Find its range and inverse function. Why does restricting the domain matter?
4
A special limit
Use a result from the supplied formulary to evaluate lim as x → 0 of (e^(4x) − 1)/(2x). State which part is printed and which part you derived.
5
Does market equilibrium exist?
Demand is D(p) = 12 − p² and supply is S(p) = 2 + 2p, for p in [0,3]. Without solving a quadratic equation, prove that exactly one market-clearing price exists in (0,3).
Reveal explained answer key
Optional deeper understanding
Beyond the 10–15-minute core lesson · Explore when you're ready for harder exercises.
A. Why oscillation doesn't always destroy continuity
Consider:
\[ f(x)= \begin{cases} x\sin(1/x),&x\neq0\\ 0,&x=0 \end{cases} \]
The expression \(\sin(1/x)\) oscillates infinitely often as \(x\to0\), but the factor \(x\) makes those oscillations shrink towards zero.
Because \(-1\leq\sin(1/x)\leq1\),
\[ |x\sin(1/x)|\leq|x| \]
And \(|x|\to0\). By the squeeze theorem:
\[ \lim_{x\to0}x\sin(1/x)=0=f(0) \]
So \(f\) is continuous at zero, despite the infinitely rapid oscillations.
The lesson: Continuity is about convergence of values, not whether a graph looks smooth.
B. Finding a root using bisection
Bolzano establishes that a root exists, but it does not calculate that root.
The bisection method uses the sign-change idea repeatedly.
Return to:
\[ f(x)=x^3+x-1 \]
We know \(f(0)<0\) and \(f(1)>0\), so a root lies in \((0,1)\).
Take the midpoint \(m=0.5\):
\[ f(0.5)=0.125+0.5-1=-0.375 \]
The function is still negative. Since \(f(1)>0\), the root must lie in \((0.5,1)\).
Repeat with the new midpoint \(0.75\):
\[ f(0.75)=0.421875+0.75-1=0.171875 \]
Now the root lies in \((0.5,0.75)\).
Each step halves the interval while preserving the sign change. This is how a theorem about existence becomes a practical numerical algorithm.
C. A harder limit with a square root
Evaluate:
\[ \lim_{x\to9}\frac{\sqrt{x}-3}{x-9} \]
Direct substitution gives \(0/0\).
Multiply numerator and denominator by the conjugate \(\sqrt{x}+3\):
\[ \begin{aligned} \frac{\sqrt{x}-3}{x-9} &=\frac{(\sqrt{x}-3)(\sqrt{x}+3)} {(x-9)(\sqrt{x}+3)}\\ &=\frac{x-9}{(x-9)(\sqrt{x}+3)}\\ &=\frac1{\sqrt{x}+3},\quad x\neq9 \end{aligned} \]
Therefore:
\[ \boxed{\lim_{x\to9}\frac{\sqrt{x}-3}{x-9}=\frac16} \]
This is a useful example of why simplification must come before evaluating an indeterminate expression.
Before you unlock Exercise Series 2
You are ready to begin if you can explain, without memorizing a template, why a function can have a limit at a missing point, why a jump cannot be repaired by changing just one value, why exponential and logarithmic functions are inverses, and why sign change plus continuity proves existence but not necessarily uniqueness.
Exercise Series 2 connects the Week 2 financial mathematics material with Week 3 functions, inverse functions, limits, and continuity. Its officially highlighted exercises include A7 (ranges and inverse functions) and A14 (continuous functions and computing limits).
Masterclass complete
Week 3
Theory prepares you to recognize the mathematical structure. The next step is to solve unfamiliar problems, inspect your reasoning, and correct mistakes.
Source references and coverage notes
This lesson is grounded in the HSG HS2026 course materials, with original Monastery explanations and examples rather than reproductions of the lecture exercises.
Official HS2026 English course schedule: MatheA-EN-HSG-information.pdf, pp. 9–10, identifies 29 September as Limits of Functions and Continuity and associates Week 3 with Exercise Series 2.
04_CURATED_STUDY_GUIDES_001.md
Official 2026 Lecture 3, German original: MatheA-DE-HSG-Vorlesung03.pdf, slides 4–25 (function foundations), 28–84 (limits, continuity and Bolzano), and the optional self-study appendix. Preserved in SUPPLEMENTARY_ORIGINALS.pdf, PDF pp. 954–1072. The examples involving rationalization, oscillations and Bolzano correspond to material covered in this lecture.
Official Mathematics A exam formulary: Canvas file 4159731, MatheA-formulary.pdf, pp. 1–2, used to verify every claim about supplied formulas.
02_CANVAS_LECTURES_AND_READINGS_001.md
Official HS2026 course and examination fact sheet: confirms functions, inverse functions, logarithms, exponentials and properties of continuous functions as examinable course topics.
03_MANUAL_NOTES_AND_READINGS_001.md
Coverage limitation: The original 2026 German Lecture 3 was available for inspection, but the corresponding English Lecture 3 slide PDF was not embedded as accessible full text in the supplied Project package. Thus, exact slide-for-slide equivalence between the language tracks remains unverified. The main Week 3 topics are supported by the official English schedule, examination fact sheet, preserved German lecture and formulary. The optional deeper material is clearly separated from the essential guided introduction.