Here are the solutions by ChatGPT to this week’s midterm. What grade would you give it?
Give your grade in the comments! (Hint: there are definitely some things which are wrong here.)
Problem 1
We need to show that
\[
\lim_{x \rightarrow 1} x^2 = 1
\]
using the \(\varepsilon\)-\(\delta\) definition of a limit.
Solution
To prove that \(\lim_{x \to 1} x^2 = 1\) using the \(\varepsilon\)-\(\delta\) definition, we must show that for any \(\varepsilon > 0\), there exists a \(\delta > 0\) such that if \(0 < |x - 1| < \delta\), then \(|x^2 - 1| < \varepsilon\).
- Starting with \(|x^2 – 1|\): We have:
\[
|x^2 – 1| = |x^2 – 1^2| = |x – 1||x + 1|.
\] - Choosing \(\delta\):
We want to make \(|x – 1||x + 1| < \varepsilon\). Notice that if \(x\) is close to 1, then \(x + 1\) will be close to 2. Let’s assume \(|x - 1| < 1\), which implies \(0 < x < 2\), so \(1 < x + 1 < 3\). - Bounding \(|x + 1|\):
In this case, \(|x + 1| < 3\). Then: \[ |x - 1||x + 1| < 3|x - 1|. \] - Setting up \(\delta\):
To ensure \(|x – 1||x + 1| < \varepsilon\), we can choose \(\delta\) such that \(3|x - 1| < \varepsilon\), or equivalently, \( |x - 1| < \frac{\varepsilon}{3} \). - Conclusion:
Let \(\delta = \frac{\varepsilon}{3}\). Then if \(0 < |x - 1| < \delta\), we have: \[ |x^2 - 1| = |x - 1||x + 1| < 3|x - 1| < 3 \cdot \frac{\varepsilon}{3} = \varepsilon. \] Thus, \(\lim_{x \to 1} x^2 = 1\) as required.
—
Problem 2
We need to construct a set \(A\) with the following properties:
- \(\sup A\) exists and is not in \(A\).
- \(\inf A\) exists and lies in \(A\).
Solution
Consider the set \(A = \{x \in \mathbb{R} : 0 \leq x < 1\}\).
- Checking \(\sup A\):
The supremum of \(A\) is 1, as 1 is the least upper bound of all elements in \(A\). However, 1 is not in \(A\) because \(A\) only contains elements less than 1. Thus, \(\sup A = 1\) exists and is not in \(A\). - Checking \(\inf A\):
The infimum of \(A\) is 0, as 0 is the greatest lower bound of all elements in \(A\). Moreover, 0 is included in \(A\), so \(\inf A = 0\) exists and lies in \(A\).
Thus, the set \(A = \{x \in \mathbb{R} : 0 \leq x < 1\}\) satisfies the required conditions.
—
Problem 3
We analyze the conditions to determine whether each implies continuity at \(x = 0\).
-
\(\exists \varepsilon > 0, \forall \delta > 0, \ \text{if \(0 < |x| < \delta\), then} \ |f(x) - f(0)| < \varepsilon.\)
This condition does not imply continuity at \(x = 0\). Here’s why:
- This condition states that there is some fixed \(\varepsilon > 0\) such that no matter how small we make \(\delta\), \(f(x)\) will always stay within \(\varepsilon\) of \(f(0)\) when \(x\) is close to 0.
- Continuity requires that for every \(\varepsilon > 0\), there is a \(\delta\) such that if \(0 < |x| < \delta\), then \(|f(x) - f(0)| < \varepsilon\).
Thus, this condition does not ensure continuity.
-
\(\forall \varepsilon > 0, \forall \delta > 0, \ \text{if \(0 < |x| < \delta\), then} \ |f(x) - f(0)| < \varepsilon.\)
This condition does imply continuity at \(x = 0\):
- This condition is actually stronger than the usual \(\varepsilon\)-\(\delta\) definition of continuity, as it requires that \(|f(x) – f(0)| < \varepsilon\) for all \(\delta > 0\) rather than just for some \(\delta\) depending on \(\varepsilon\).
- Therefore, this condition guarantees continuity at \(x = 0\).
-
\(\forall \varepsilon > 0, \exists \delta > 0, \ \text{if \(0 < |x| < \delta\), then} \ |f(x) - f(0)| < \varepsilon^2.\)
This condition does not imply continuity at \(x = 0\):
- This condition is weaker than the standard \(\varepsilon\)-\(\delta\) definition because it bounds \(|f(x) – f(0)|\) by \(\varepsilon^2\) instead of \(\varepsilon\).
- The fact that the bound involves \(\varepsilon^2\) instead of \(\varepsilon\) means it could still be large enough that \(|f(x) – f(0)|\) doesn’t approach 0 as \(\varepsilon \to 0\).
Therefore, this does not imply continuity at \(x = 0\).
—
Problem 4
Let \(f(x)\) be a continuous function on \(\mathbb{R}\) satisfying \(f(x+1) = f(x)\) for all \(x \in \mathbb{R}\). We need to prove that \(f(x)\) is bounded on \(\mathbb{R}\).
Solution
- Periodicity: The condition \(f(x+1) = f(x)\) implies that \(f(x)\) is periodic with period 1. This means that the behavior of \(f(x)\) on the entire real line is determined by its behavior on any interval of length 1, for instance, the interval \([0, 1]\).
- Continuity on \([0, 1]\): Since \(f\) is continuous on \(\mathbb{R}\), it is also continuous on the closed interval \([0, 1]\).
- Boundedness on \([0, 1]\): By the Extreme Value Theorem, a continuous function on a closed interval \([0, 1]\) attains its maximum and minimum values on that interval. Thus, there exist constants \(m\) and \(M\) such that \(m \leq f(x) \leq M\) for all \(x \in [0, 1]\).
- Extending the Bound: For any \(x \in \mathbb{R}\), there exists an integer \(k\) such that \(x – k \in [0, 1]\). Using the periodicity, we have:
\[
f(x) = f(x – k + k) = f(x – k).
\]
Since \(x – k \in [0, 1]\), it follows that \(m \leq f(x) \leq M\) for all \(x \in \mathbb{R}\).
Therefore, \(f(x)\) is bounded on \(\mathbb{R}\).