calculus - Explanation for the convergence of the limits

I have the limits and I am wondering why it convergs?

$$\lim_{x \rightarrow \infty} e^{-x} \Big(x^3 -3x \Big)$$
also it converges to $0$ although it equals $(0 \times \infty) \neq 0$

Example of real analytic function

We were taught real analytic functions in class today. I am playing around trying to construct examples. I see exponential, sine, cosine and logarithmic functions (for $x > 0$). One function I am having trouble with is $f(x) = \frac{1}{1 + e^x}$. In spirit, this function is like $e^{-x}$, so I want to say it is real analytic, but not totally sure. Any help, please?

Answer

To recall that the reciprocal of an analytic function with no zeros is analytic is one way. For an argument see Is the reciprocal of an analytic function analytic?

elementary set theory - Can $bigcap_{Bin A}B=emptyset$ Given that all elements of $A$ are inductive sets?

I am reading a course in mathematical analysis vol 1 by J.H. Garling.

He defines a successor set as one that (1) contains $\emptyset$, and (2) contains $a^+$ whenever it contains $a$ (where $a^+$ is defined as the set $a\cup \{a\}$. He then states as a Theorem that there exists a successor set $Z^+$ such that any successor set $T$ must contain $Z^+$. Unfortunately I cannot understand the proof which is as follows:

"Note that if $A$ is a set, all of whose elements are successor sets, then it follows immediately from the definitions that the intersection of all elements of $A$ is also a successor set. Suppose that $S$ is a successor set. Let $Z^+=\cap\{B\in P(S):B\text{ is a successor set}\}$. Then if $T$ is a successor set, $T\cap S$ is a successor set, so $Z^+\subseteq T\cap S\subseteq T$."

In particular, I do not see why it is obvious that the intersection of successor sets is always a successor set, I see why that is the case if the intersection is non-empty, but i can't see why it couldn't be the empty set itself.

Answer

Especially the first part of this answers your question. I read in the comments that there were more difficulties so decided to give a more complete answer.

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Let $S$ be a successor-set.

If $\mathcal A:=\{B\in\wp(S)\mid B\text{ is a successor set}\}$ then $S\in\mathcal A$.

This guarantees that $\cap\mathcal A$ is a well defined subset of $S$.

Secondly every element of $\mathcal A$ is a successor set so that $\varnothing\in B$ is true for every $B\in\mathcal A$.

Consequently $\varnothing\in\cap\mathcal A$ showing that the intersection cannot be empty.

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Let $\omega:=\cap\mathcal A$. We will prove that $\omega$ is a successor set.

As stated above we have $\varnothing\in\omega$. If $a\in\omega$ then for every $B\in\mathcal A$ we have $a\in B$. Every $B\in\mathcal A$ is a successor set, so we conclude that $a^+\in B$ for every $B\in\mathcal A$. That means exactly that $a^+\in\omega$ and proved is now that $\omega$ is a successor set.

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Here a proof that $\omega\subseteq T$ is true for every $T$ that is a successor set.

Let $T$ be a successor set. Then $T\cap S\in\wp(S)$ is a successor set so that $T\cap S\in\mathcal A$.

Then $\omega\subseteq T\cap S\subseteq T$.

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Final remark:

Essential is here the existence of a successor set $S$. Without that the reasoning could not have been made. The statement that a successor set exists is the so-called axiom of infinity.

calculus - Does, $mathop {lim }limits_{x to +infty } f'(x) = + infty Leftrightarrow mathop {lim }limits_{x to +infty } frac{{f(x)}}{x} = + infty $?

Let $f:\Bbb R \to \Bbb R$ be a differentiable function. If $\mathop {\lim }\limits_{x \to + \infty } \frac{{f(x)}}{x} = + \infty $, it is always true that $\mathop {\lim }\limits_{x \to + \infty } f'(x) = + \infty $? How about the converse?

For example, $\mathop {\lim }\limits_{x \to + \infty } \frac{{\ln x}}{x} = 0$ is finite, then we can see $\mathop {\lim }\limits_{x \to + \infty } (\ln x)' = 0$ is finite. $\mathop {\lim }\limits_{x \to + \infty } \frac{{{x^2}}}{x} = + \infty $ so $\mathop {\lim }\limits_{x \to + \infty } ({x^2})' = \mathop {\lim }\limits_{x \to + \infty } x = + \infty $. So the claim seems good to me, but I don't know how to actually prove it. $\mathop {\lim }\limits_{x \to + \infty } f'(x) = \mathop {\lim }\limits_{x \to \infty } \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}$, I don't know how to deal with this mixed limit. Also since the limits in the proposition diverges, it looks like mean value theorem sort of thing cannot apply here.

Answer

The "left to right" of the biconditional is true. As noted in another answer, we can use L'hopital. But I will utilize a direct approach. We need to show that for arbitrarily large $M$, we have for sufficiently large $x$ the inequality $\frac{f(x)}{x} > M$.

By assumption, for any arbitrarily large $M$ we have $f'(x) > 2M$ when $x>x_0$. This means $f(x) \geq f(x_0) + 2M(x-x_0)$ for $x > x_{0}$. Note also that there is an $x_1$ such that for all $x > x_1$, $f(x_0) + 2M(x-x_0) > {Mx}$. Hence, we can see that for $x > \max\{x_1, x_0\}$ we have $$\frac{f(x)}{x} \geq \frac{f(x_0) + 2M(x-x_0)}{x} > \frac{Mx}{x} = M $$

The "right to left" of the biconditional is false. Consider $f(x) = x^2(\sin x + 2)$. This is positive and bounded below by $x^2$, hence $\lim_{x \to +\infty} \frac{f(x)}{x} = +\infty$ but $f'$ oscillates as $x \to +\infty$.

We can say something weaker, however, namely the following

Theorem: Let $f \in C^1(\mathbb{R})$ such that $$\lim_{x \to +\infty} \frac{f(x)}{x} = +\infty$$ Then we have $$\limsup_{x \to +\infty} \ f'(x) = +\infty$$

To prove this, first note that for $f$, we can assume $f(0) = 0$ without any loss of generality. Indeed, define $g(x) = f(x) - f(0)$ and note $\lim_{x \to +\infty} \frac{f(x)}{x} = +\infty \Longleftrightarrow \lim_{x \to +\infty} \frac{g(x)}{x} = +\infty$ and also $f' = g'$.

We can prove by contradiction. Suppose the $\lim \sup$ is finite or $-\infty$. This means $f'$ is bounded above in $[M, +\infty)$ for some $M>0$. Since $f'$ is continuous, by the extreme value theorem it is bounded above in $[0,M]$, and hence it is bounded above in $[0, +\infty)$. By the mean value theorem, we have that $\frac{f(x)}{x} = f'(\alpha)$ for some $\alpha$ in $[0, x]$. Letting $x \to +\infty$ we can see that $f'(\alpha)$ takes on arbitrarily large positive values, which contradicts the fact that $f'$ is bounded above in $[0, +\infty)$.

This can probably be modified so that the $C^1$ condition can be relaxed (e.g., to allow for cases where $f'$ is discontinuous), but I'm not sure how to do that.

analysis - Inequality for $|f(x_n)-f(x_n+(y_n-x_n)|$.

If $f: \mathbb{R} \to \mathbb{R}$ is a function and $a_n$ a sequence of real numbers.We have $f_{a_n(x)}=f(x+a_n)$.Show that if for any zero sequence $a_n$ we have $f_{a_n} \to f$ uniformly,then $f$ is uniformly continuous.

To show that $f$ is uniformly continuous,it suffives to show that if $x_n,y_n$ are two sequences of real numbers and $x_n-y_n \to 0$,then $f(x_n)-f(y_n) \to 0$.

$|f(x_n)-f(y_n)|=|f(x_n)-f(x_n+(y_n-x_n)| \leq $ what?

We know that $f_{a_n} \to f$ uniformly, that means that $\exists n_0$ such that $\forall n \geq n_0: |f_{a_n}-f|_{\mathbb{R}}= \sup_{x \in \mathbb{R}} |f_{a_n}-f| \leq \epsilon$

How can I find an inequality for $|f(x_n)-f(x_n+(y_n-x_n)|$ ?

Answer

You've basically got it. Just let $a_n=y_n-x_n$. This is your zero sequence (which I assume means a sequence that converges to $0$). Then

$$|f(x_n)-f(y_n)|=|f(x_n)-f(x_n+a_n)=|f(x_n)-f_{a_n}(x_n)|\le\sup_{x\in\mathbb{R}}|f(x)-f_{a_n}(x)|\le\epsilon$$

linear algebra - How to find the correct order to multiply elimination matrices?

Let's say I have the following matrix $A \in \mathbb{R}^{4\times4}$

$$A = \begin{bmatrix}

a & a & a & a \\
a & b & b & b \\
a & b & c & c \\
a & b & c & d
\end{bmatrix}$$

And I perform the following Row Operations (which are really just matrix multiplications) to reduce $A$ into an upper triangular matrix $U$

1. $R_2 - (l_{21} = 1)R_1$ ($\text{corresponds to }E_{21})$



2. $R_3 - (l_{31} = 1)R_2$ ($\text{corresponds to }E_{31})$


3. $R_4 - (l_{41} = 1)R_3$ ($\text{corresponds to }E_{41})$


4. $R_3 - (l_{32} = 1)R_2$ ($\text{corresponds to }E_{32})$


5. $R_4 - (l_{42} = 1)R_3$ ($\text{corresponds to }E_{42})$


6. $R_4 - (l_{43} = \frac{c-a}{c-b})R_3$ ($\text{corresponds to }E_{43})$

After performing all these row operations we arrive at
$$U = \begin{bmatrix}
a & a & a & a \\

0 & b-a & b-a & b-a \\
0 & 0 & c-b & c-b \\
0 & 0 & 0 & d-c
\end{bmatrix}$$

But now if I want to find the elimination matrix $E$, that does all of these row operation in one step? How do I go about finding it? To re-iterate, I'm trying to find a $E$ such that $EA = U$.

I understand $E$ would be the product of the elimination matrices used in the row operations, but in what order would the matrices be multiplied? The order being of importance as matrix multiplication isn't commutative.

As it turns out, $E \neq E_{21}E_{31}E_{41}E_{32}E_{42}E_{43}$

Is there a theorem/general rule that can be used to find the correct order to multiply the matrices used in the row operations, so that a single elimination matrix can be found without trying every possible permutation?

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A Second Example, $B \in \mathbb{R}^{3\times3}$

$$B = \begin{bmatrix}
1 & 4 & 0 \\
4 & 12 & 4 \\

0 & 4 & 0 \\
\end{bmatrix}$$

The following row operations are performed to transform $B$ into an upper triangular $U$

1. $R_2 - (l_{21} = 4)R_1$ $\text{(corresponds to } E_{21})$


2. $R_3 - (l_{32} = -1)R_2$ $\text{(corresponds to } E_{32})$

We arrive at

$$U = \begin{bmatrix}
1 & 4 & 0 \\
0 & 1 & -1 \\
0 & 0 & 1 \\
\end{bmatrix}$$

In this case, $E_{32}E_{21}B \neq U$, however $E_{21}E_{32}B = U$, going against the convention in example 1

Answer

Perhaps I am misunderstanding something, but here is the answer I think you're looking for. Suppose that we first conduct the row operation ${\bf R}_1$ on the matrix ${\bf A}$, then the resulting matrix is ${\bf A}_1 = {\bf R}_1 {\bf A}$. The next row operation ${\bf R}_2$ is then applied to ${\bf A}_1$, resulting in ${\bf A}_2 = {\bf R}_2 {\bf A}_1 = {\bf R}_2 {\bf R}_1 {\bf A}$. Continuing this way, you can see that if you apply $n$ row operations ${\bf R}_1, \ldots ,{\bf R}_n$, the result will be ${\bf A}_n = {\bf R}_n {\bf R}_{n-1} \cdots {\bf R}_1 {\bf A}$, showing you the order in which to multiply the operations ${\bf R}_i$.

calculus - Spot mistake in finding $lim limits_{xto1}left(frac x {x-1} - frac1 {log(x)} right)$

This is the limit I'm trying to solve: $\lim \limits_{x\to1}\left(\frac
x {x-1} - \frac1 {\log(x)}
\right)$

I thought: let's define $x=k+1$, so that $k\to0$ as $x\to1$.

Then it becomes:
$$\lim \limits_{k\to0}\left(\frac
{k+1} {k} - \frac1 {\log(k+1)}
\right)$$
and then,
$$\lim \limits_{k\to0}\left(\frac
{k+1} {k} - \frac1 {\frac {\log(k+1)\times k}k}

\right)=\lim \limits_{k\to0}\left(\frac
{k+1} {k} - \frac1 {k}
\right).$$
Which results in $\frac k k$ , that should be 1, but wolfram says it's $\frac 1 2$...

Did I do something illegal?

Answer

Yes, the illegal part is this step:
$$\lim_{k \to 0}\frac1{\frac{\log(k + 1)}k}\frac1k = \lim_{k \to 0}\frac1k$$

I see that you applied the known limit
$$\lim_{t \to 0}\frac{\log(1 + t)}t = 1$$
but the fact is that
$$\lim_{x \to \alpha}f(x)g(x) = \lim_{x \to \alpha}f(x)\times\lim_{x \to \alpha}g(x)$$
is only valid when both limit are finite. In your case you're left with $\lim\limits_{k \to 0}\frac1k$, which not only is not finite, but does not exist entirely.

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If you are looking for a way to evaluate your limit, I'd suggest MacLauring (that is, a Taylor expansion around $x =0$), which is the simplest and most elegant way. But since you said that you can't use Taylor yet, I fear your only possibility is going with L'Hospital.