Abstract
Quiz 4 in the course Infinitesimal Calculus (1), semester 2026 A.
Quiz instructions: The duration of the quiz is 36 minutes.
The use of calculators and of any other auxiliary material is forbidden.
This is a quiz that I translated and solved.
The original quiz was written by the Infi Calculus 1 course lecturers at HUJI.
Problem (Question 1 - 65 points). Prove or disprove:
If \(f:\mathbb{R}\to\mathbb{R}\) is a continuous function at \(x_{0}\in\mathbb{R}\) and in every punctured neighborhood of \(x_{0}\) there exists a point \(z\in\mathbb{R}\) s.t. \(f(z)=z^{2}\), then \(f(x_{0})=x_{0}^{2}\).
Proof. using the \(\varepsilon-\delta\) definition:
1. Since \(f\) is continuous at \(x_{0}\), \[\forall\varepsilon>0\exists\delta>0\ \text{s.t.}\ \forall x\in\left(x_{0}-\delta,x_{0}+\delta\right):\left|f(x)-f(x_{0})\right|<\varepsilon\]
2. We shall define \(g:\mathbb{R}\rightarrow\mathbb{R}\) by \(g(x)=x^{2}\).
The function \(g\) is continuous on \(\mathbb{R}\) because it is a polynomial, hence the same continuity characterisation as in point 1 applies to it.
3. Let us assume for the sake of contradiction that \(f(x_{0})\ne x_{0}^{2}\). Immediately, \(f(x_{0})\ne g(x_{0})\).
4. The law of trichotomy establishes that either \(f(x_{0})>g(x_{0})\) or \(f(x_{0})<g(x_{0})\).
We will assume without loss of generality that \(f(x_{0})>g(x_{0})\).
Hence, \(f(x_{0})-g(x_{0})>0\).
5. The previous point allows us to apply the known technique of halving the distance to separate these values and set \(\varepsilon:=\frac{f(x_{0})-g(x_{0})}{2}>0\) in the property of points 1 and 2, which gives us corresponding deltas \(\delta_{1},\delta_{2}>0\).
We shall define \(\delta_{3}:=\min\left\{ \delta_{1},\delta_{2}\right\}\) and get the combined property: \[\forall x\in\left(x_{0}-\delta_{3},x_{0}+\delta_{3}\right)\setminus\left\{ x_{0}\right\} :\left|f(x)-f(x_{0})\right|<\frac{f(x_{0})-g(x_{0})}{2}\ \wedge\left|g(x)-g(x_{0})\right|<\frac{f(x_{0})-g(x_{0})}{2}\]
Expanding the absolute values gives: \[\frac{g(x_{0})-f(x_{0})}{2}<f(x)-f(x_{0})<\frac{f(x_{0})-g(x_{0})}{2}\ \wedge\ \frac{g(x_{0})-f(x_{0})}{2}<g(x)-g(x_{0})<\frac{f(x_{0})-g(x_{0})}{2}\]
Isolating \(f(x)\) and \(g(x)\) on the relevant sides provides a lower bound for \(f(x)\) and an upper bound for \(g(x)\): \[\frac{g(x_{0})+f(x_{0})}{2}<f(x)\quad\wedge\quad g(x)<\frac{f(x_{0})+g(x_{0})}{2}\]
Thus, in particular, \[\forall x\in\left(x_{0}-\delta_{3},x_{0}+\delta_{3}\right)\setminus\left\{ x_{0}\right\} :g(x)<f(x)\]
6. This means that \[\forall x\in\left(x_{0}-\delta_{3},x_{0}+\delta_{3}\right)\setminus\left\{ x_{0}\right\} :f(x)\ne x^{2}\] In other words, \(\left(x_{0}-\delta_{3},x_{0}+\delta_{3}\right)\setminus\left\{ x_{0}\right\}\) is a punctured neighborhood of \(x_{0}\) which contradicts the assumption that every punctured neighborhood of \(x_{0}\) contains a point \(z\in\mathbb{R}\) s.t.\(f(z)=z^{2}\). ∎
Proof. using Heine’s sequences:
1. By the assumption, for each \(n\in\mathbb{N}\), there exists a point \[x_{n}\in\left(x_{0}-\frac{1}{n},x_{0}+\frac{1}{n}\right)\setminus\left\{ x_{0}\right\} \quad\text{s.t.\ensuremath{\quad}}f(x_{n})=x_{n}^{2}\] Thus, we have a sequence \(\left(x_{n}\right)_{n=1}^{\infty}\) such that for all \(n\in\mathbb{N},\) \[x_{0}-\frac{1}{n}<x_{n}<x_{0}+\frac{1}{n}\ \wedge\ x_{n}\ne x_{0}\ \wedge\ f(x_{n})=x_{n}^{2}\]
2. Since \(\underset{n\rightarrow\infty}{\lim}\left(x_{0}-\frac{1}{n}\right)=\underset{n\rightarrow\infty}{\lim}\left(x_{0}+\frac{1}{n}\right)=x_{0}\), the squeeze theorem implies \(\underset{n\rightarrow\infty}{\lim}x_{n}=x_{0}\).
3. Points 1 and 2 show that \(\left(x_{n}\right)_{n=1}^{\infty}\) satisfies the conditions of a Heine sequence for continuity.
By the assumption, \(f\) is continuous at \(x_{0}\), therefore \(\underset{n\rightarrow\infty}{\lim}f(x_{n})=f(x_{0})\).
4. According to the arithmetic of sequence limits, \[\underset{n\rightarrow\infty}{\lim}f(x_{n})=\underset{n\rightarrow\infty}{\lim}x_{n}^{2}=\underset{n\rightarrow\infty}{\lim}x_{n}\cdot\underset{n\rightarrow\infty}{\lim}x_{n}=x_{0}\cdot x_{0}=x_{0}^{2}\] 5. Therefore, \(f(x_{0})=x_{0}^{2}\), as required. ∎
Problem (Question 2 - 35 points). Let \((a_{n})_{n=1}^{\infty}\) be a sequence of natural numbers that satisfies \(\underset{n\rightarrow\infty}{\lim}\frac{a_{n}}{n}=\infty\).
Prove that \(\underset{n\rightarrow\infty}{\lim}\left(1+\frac{1}{n}\right)^{a_{n}}=\infty\).
Proof. Since \(\underset{n\rightarrow\infty}{\lim}\frac{a_{n}}{n}=\infty\), the arithmetic of sequence limits ensures \(\underset{n\rightarrow\infty}{\lim}\left(1+\frac{a_{n}}{n}\right)=\infty\).
Additionally, because \(a_{n}\in\mathbb{N}\) and \(\frac{1}{n}>-1\) for all \(n\in\mathbb{N}\), Bernoulli’s inequality holds and \[\forall n\in\mathbb{N}:\left(1+\frac{1}{n}\right)^{a_{n}}\ge1+\frac{a_{n}}{n}\]
Recall the slice theorem: for two sequences \((b_{n})_{n=1}^{\infty}\) and \((c_{n})_{n=1}^{\infty}\), if there exists \(N\in\mathbb{N}\) such that for every \(n>N\) the inequality \(b_{n}\ge c_{n}\) holds, and \(\underset{n\rightarrow\infty}{\lim}c_{n}=\infty\), then \(\underset{n\rightarrow\infty}{\lim}b_{n}=\infty\).
Therefore, \(\underset{n\rightarrow\infty}{\lim}\left(1+\frac{1}{n}\right)^{a_{n}}=\infty\) as required. ∎