Mid
SLCM_Exam_Notebook_999027873_202401-104195-5_Datestamp_2024_12_11.pdf
4(c) i have no problem of correction of 4(c). Here is just my correction to my mistake
\(a_{n}=\left(-1\right)^{n}\frac{1}{\sqrt{n}}\)
\(\sum|a_{n+1}-a_{n}|=\sum\left|\frac{1}{\sqrt{n+1}}+\frac{1}{\sqrt{n}}\right|=\sum\left|\frac{\sqrt{n}+\sqrt{n+1}}{\sqrt{n+1}\sqrt{n}}\right|=\sum\frac{1}{\sqrt{n+1}\sqrt{n}\left(\sqrt{n+1}-\sqrt{n}\right)}=\sum\frac{1}{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}\)
And \(\sum\frac{1}{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}\geq\sum\frac{1}{\left(n+1\right)\sqrt{n}-n\sqrt{n}}=\sum\frac{1}{\sqrt{n}}\)
Or
4(c) \(a_{n}=\left(-1\right)^{n}\frac{1}{n}\)
\(\sum|a_{n+1}-a_{n}|=\sum\left|\frac{1}{n+1}+\frac{1}{n}\right|=\sum\frac{1}{n+1}+\sum\frac{1}{n}\) which is divergent
3(b) 6points i have problem of correction of 3(b).
Here i don't understand why it is not clear enough. At the last part of proof, since \(-\epsilon_1 - \epsilon_2 < 0\), then \(-\epsilon_1 - \epsilon_2\) is a negative number. And we know \(a-b>-\epsilon_1-\epsilon_2\), thus \(a-b>\) negative number, thus it is greater or equal to \(0\). I think it's ok
Here is the original proof in exam without modification
Since \(a_n = \sup\{x_k: k \leq n\}\), then \(a_n \geq x_k\) for \(k \leq n\).
Since \(b_n = \inf\{x_k: k \geq n\}\), then \(b_n \leq x_k\) for \(k \geq n\).
Thus \(b_n \leq x_k \leq a_n\) for \(b_n\) such that \(k \geq n\).
Thus \(a_n \geq b_n\).\(\forall n\)
Since \(a = \lim a_n\), then \(|a_n - a| < \epsilon_1\) (\(\forall \epsilon_1 > 0\), \(\exists N\))
Since \(b = \lim b_n\), then \(|b_n - b| < \epsilon_2\).(\(\forall\epsilon_2>0\), \(\exists N\)))
Then \(-\epsilon_1 + a < a_n < \epsilon_1 + a\), and \(-\epsilon_2 + b < b_n < \epsilon_2 + b\).
Also since \(a_n \geq b_n\), then \(a+\varepsilon_1\geq a_{n}\geq b_{n}\geq b-\epsilon_2\).
Thus we have \(a+\epsilon_1\geq b-\epsilon_2\Rightarrow a-b>-\epsilon_1-\epsilon_2\).
Since \(\epsilon_1, \epsilon_2 > 0\), then \(-\epsilon_1 - \epsilon_2 < 0\).
Then \(a - b\) is bigger than a negative number.
Then \(a - b \geq 0 \Rightarrow a \geq b\).
3(c) total 4 points i have problem of correction of 3(c).
Proof
\(\Leftarrow\)) We know If \(a = b \Rightarrow \lim_{n \to \infty} a_n = \lim_{n \to \infty} b_n\), then \(\limsup_{n\to\infty}\{x_{k}:k\geq n\}=\liminf_{n\to\infty}\{x_{k}:k\geq n\}\)(*)
Suppose \((x_n)\) is not convergent, then \(\exists\varepsilon>0\,,\forall n>n_0:|x_{n}-L|\geq\varepsilon\).
Then \(x_{n}\geq L+\varepsilon\quad\text{or}\quad x_{n}\leq L-\varepsilon\quad\)\(\forall n>n_0\)
Then \(\inf\{x_{n}:n\geq n_0\}\geq L+\varepsilon\quad\text{or}\quad\sup\{x_{n}:n\geq n_0\}\leq L-\varepsilon\)
\(\Rightarrow\inf\{x_{k}:k\geq n\}\geq L+\varepsilon\quad\text{or}\quad\sup\{x_{k}:k\geq n\}\leq L-\varepsilon\)
\(\Rightarrow\liminf_{n\to\infty}\{x_{k}:k\geq n\}\geq L+\varepsilon\quad\text{or}\quad\limsup_{n\to\infty}\{x_{k}:k\geq n\}\leq L-\varepsilon\)
Since(*) we know \(L + \varepsilon \leq L - \varepsilon \Rightarrow 2\varepsilon \leq 0\), contradiction!
Thus \((x_n)\) is convergent.
Above is one direction of original proof in exam, i think it is clear, but the grader says it is not clear enough.
\(\Rightarrow\)) Suppose \(a \neq b\), since \(a \geq b \Rightarrow a > b\).
Then \(\lim_{n\to\infty}a_{n}>\lim_{n\to\infty}b_{n}\Rightarrow\limsup_{n\to\infty}\{x_{k}:k\geq n\}>\liminf_{n\to\infty}\{x_{k}:k\geq n\}\).
\(\Rightarrow\exists x_{k}\,:k\geq n\,\text{such that}\,\inf\{x_{k}:k\geq n\}<x_{k}<\sup\{x_{k}:k\geq n\}\) which means there is always \(x_{k}\,(k\geq n)\) when \(n \to \infty\) in the interval.
And \(\inf\{x_{k}:k\geq n\},\sup\{x_{k}:k\geq n\}\) also in the set.
Thus when \(n \to \infty\), there always at least three elements satisfying condition.
Thus there is no limit. Contradiction!
Finally, we complete proof.
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(20 points) Let \(F : \mathbb{R} \to \mathbb{R}\) be the function defined by
\[ f(x) = \begin{cases} x, & \text{if } x \in \mathbb{Q}; \\ 0, & \text{if } x \notin \mathbb{Q}. \end{cases} \]Prove that \(\lim_{x \to 0} f(x) = 0\) and that \(\lim_{x \to a} f(x)\) does not exist for all \(a \neq 0\).
Proof
To prove \(\lim_{x \to a} f(x)\) does not exist for all \(a \neq 0\). Use contradiction. Suppose \(\lim_{x \to a} f(x)\) exists for some \(a\neq0\)
Then \(\lim_{x\to a}f(x)=l\)
Thus \(\forall\varepsilon>0,\exists\delta>0:0<|x-a|<\delta\Rightarrow|f(x)-l|<\varepsilon\).
Then \(0<\left|x-a\right|<\delta\Rightarrow l-\varepsilon<f(x)<l+\varepsilon\).
By density, we can take \(x=q\in \mathbb{Q}\) which is rational number, then \(f(x)=x\Rightarrow l-\varepsilon<x<l+\varepsilon\)
We also can take \(x=i\in \mathbb{I}\) which is irrational number, then \(f(x)=0\Rightarrow l-\varepsilon<0<l+\varepsilon\)
\(|x|<2\varepsilon\)
or take a rational subsequence and irrational subsequence
since the limit of function equal to the limit of sequence equal to the limit of subsequence
