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14

It's possible that your teacher was pointing out the fact that the function doesn't exist at $x = 1$. That's different from saying that the limit doesn't exist as $x \to 1$. Notice that by factoring, $$ f(x) = \frac{x-1}{x^2 + 2x - 3} = \frac{x-1}{(x-1)(x+3)} $$ As long as we are considering $x \ne 1$, the last expression simplifies: $$ ...


9

Your teacher is not correct. There are two easy ways to do this problem. The first is by factoring the denomiator: $$\lim_{x\to 1}\frac{x-1}{(x-1)(x+3)}=\lim_{x\to 1}\frac{1}{x+3}=\frac{1}{4}$$ The second is by using L'Hospital's rule, which is a useful identity in limits. By L'Hospital's rule, we know that $$\lim_{x\to 1}\frac{x-1}{x^2+2x-3}=\lim_{x\to ...


8

Note that if $f(x)=0$ for some $x$, then $f$ is constantly $0$, since $f(z)=f\left(x+\left(z-x\right)\right)$. So proceed assuming that $f$ never outputs $0$. We have $$ \begin{align} f\left(\left(x+y\right)+z\right)&=f(x+y)\, f(z)\, f(xz+yz)\\ &=f(x)\,f(y)\,f(xy)\,f(z)\,f(xz)\,f(yz)\,f(xyz^2) \end{align}$$ But also $$ \begin{align} ...


8

If $f$ is continuous, then the answer is: Yes. If $f(x_1)=y_1<y_2=f(x_2)$, then $[y_1,y_2]\subset f[x_1,x_2]$, due to Intermediate Value Theorem. However, if we allow $f$ to be discontinuous, then the answer is: No. There exists an 1-1 and onto mapping between $[0,1]$ and the Cantor set, which is nowhere dense, and is equinumerous to $[0,1]$.


6

For the limit as $x\to\infty$, divide top and bottom by $e^{2x}$. For the limit as $x\to-\infty$, it is enough to look.


6

I don't know what your teacher means exactly. Limits are defined when $x$ tends to some number, or infinity. Not when $x$ is this number. The value that the function takes at the limit poit is irrelevant (to compute the limit). In fact, in most high school limit exercises, the function is not defined at the point that $x$ tends to.


4

What about a little l'Hospital? $$\lim_{x\to 1}\frac{x^{1/n}-1}{x^{1/m}-1}=\lim_{x\to 1}\frac{\frac1nx^{1/n -1}}{\frac1mx^{1/m-1}}=\lim_{x\to 1}\frac mn x^{\frac1n-\frac1m}=\frac mn$$


3

Note that, as $x\to 1$, $$ \frac{\sqrt[n]{x}-1}{\sqrt[m]{x}-1}=\frac{\frac{\sqrt[n]{x}-1}{x-1}}{\frac{\sqrt[m]{x}-1}{x-1}}=\frac{\frac{f(x)-f(1)}{x-1}}{\frac{g(x)-g(1)}{x-1}} \longrightarrow\frac{f'(1)}{g'(1)}, $$ where $$ f(x)=\sqrt[n]{x}, \quad g(x)=\sqrt[m]{x}. $$ And as $$ f'(x)=\big(\sqrt[n]{x}\big)'=\frac{1}{nx^{1-1/n}}, \quad\text{and}\quad ...


3

It’s not possible. Suppose that $f:[a,b]\to(a,b)$ is continuous except at finitely many points. Let the points of discontinuity in $(a,b)$ be $x_1<\ldots<x_{n-1}$. (Note that $f$ may also be discontinuous at one or both endpoints.) Let $x_0=a$ and $x_n=b$. Then $f$ is continuous on $I_k=(x_k,x_{k+1})$ for $k=0,\ldots,n-1$. The sets $f[I_k]$ for ...


3

In abstract category theory morphisms are not required to be functions: we are given an arbitrary class, regarding its elements as arrows, equipped with domain and codomain information and an associative operation. If we want, we can specify so that for a funtion $f:A\to B$, let ${\rm dom\,}f:=B$ and ${\rm cod\,}f:=A$ and the operation is defined as ...


3

Hint : What happens if $x=0$ ?


2

For any $I \subset \mathbb{N}$, define $f_I : \mathbb{N}\to \mathbb{R}$ by $f_I(n)=1$ if $n \in I$ and $0$ otherwise. You can show that $\{ f_I \mid \emptyset \neq I \subset \mathbb{N}\}$ is a family of linearly independent functions of cardinality $|\mathfrak{P}(\mathbb{N})|= |\mathbb{R}|$. Therefore, the dimension of $V$ is uncountable whereas $\dim(W)$ is ...


2

If you look back at the origin of the problem, this function is a bijection from $\mathbb N \times \mathbb N$ to $\mathbb N$. To find the inverse, start by finding a $k$ so that $k^2 \leq z < (k+1)^2$. Than see what happens depending if $z-k^2$ is larger or smaller than $k$.


2

Yes, in fact, every $C^1$ function is locally Lipchitz. This is because there is an interval around every $x$ where $f'(x)$ exists and is bounded; the bound on $f'(x)$ will be the Lipchitz constant in this neighborhood.


2

After expanding everything out, we want to factor out an $h$ so that the $h$'s cancel: \begin{align*} (x + h)^3 - x^3 &= (x^3 + 3hx^2 + 3h^2x + h^3) - x^3 \\ &= 3hx^2 + 3h^2x + h^3 \\ &= h(3x^2 + 3hx + h^2) \end{align*}


2

For (b) first note that $f_{V^0}+f_{W^0}\in V^0+W^0$ implies $$ (f_{V^0}+f_{W^0})(s)=f_{V^0}(s)+f_{W^0}(s)=0+0=0 $$ whenever $s\in W\cap V$ since $W\cap V$ is a subspace of both $V$ and $W$. This proves that $V^0+W^0\subset(V\cap W)^0$ Now, note that \begin{align*} \dim\bigl((V\cap W)^0\bigr) &= \dim(U)-\dim(V\cap W) \\ &= ...


2

In general, for closed $K$, the set $f(K\times K)$ is not closed. A simple example: $$K = \{ n + 2^{-(n+3)} : n \in \mathbb{N}\}.$$ Then $1 \in \overline{f(K\times K)} \setminus K\times K$.


2

In the second limit you can just use the fact that $e^{-\infty}=0$ and get the limit. In the first limit we can write $\displaystyle \frac{e^{2x}-1}{e^{2x}+1}=\frac{1-\frac{1}{e^{2x}}}{1+\frac{1}{e^{2x}}}$, hence $$\lim_{x\to\infty}\frac{e^{2x}-1}{e^{2x}+1}=\lim_{x\to\infty}\frac{1-\frac{1}{e^{2x}}}{1+\frac{1}{e^{2x}}}=\frac{1}{1}=1$$


2

l'Hopital's rule is for computing the limit of a ratio when the numerator and denominator both approach $\infty$ or $0$ (this is called an indeterminant form). Often after an application of l'Hopital's rule, a simplification of the resulting function will be necessary. Other times, the resulting limit will also be in this indeterminant form. Specifically ...


2

Partal solution: The case of $m=n-1$ can be solved easily. It is easy to check that $$ D^{(n,n-1)} = \sum_{1\le i < j \le n} \frac{p_1+\cdots + p_{i-1}+p_{i+1}+\cdots+p_{j-1}+p_{j+1}+\cdots +p_n}{2p_1+\cdots + 2p_{i-1}+p_i+2p_{i+1}+\cdots+2p_{j-1}+p_j+2p_{j+1}+\cdots +2p_n} \\ =\sum_{1\le i < j \le n} \frac{1-p_i-p_j}{2-p_i-p_j}$$ Now since ...


2

Hint: $\frac{x^{-0.5}}{x^{0.5}}=x^{-0.5-0.5}=x^{-1}=\frac{1}{x}$ and $\frac{y^{0.5}}{y^{-0.5}}=y^{0.5-(-0.5)}=y^{0.5+0.5}=y^1=y$


2

You may write $$\frac{\sqrt[n]{x}-1}{\sqrt[m]{x}-1}=\frac{e^{\frac{\ln x}{n}}-1}{e^{\frac{\ln x}{m}}-1}$$ then, as $x$ tends to $1$, you have $$e^{\frac{\ln x}{n}}= 1+\frac{\ln x}{n}+o\left(\frac{\ln x}{n}\right)$$ $$e^{\frac{\ln x}{m}}= 1+\frac{\ln x}{m}+o\left(\frac{\ln x}{m}\right)$$ finally you get $$ \frac{\frac{\ln x}{n}+o\left(\frac{\ln ...


2

We assume that $m$ and $n$ are positive integers. Let $x=y^{mn}$. We are looking for $$\lim_{y\to 1} \frac{y^m-1}{y^n-1}.\tag{1}$$ Dividing top and bottom by $y-1$, we find that the limit (1) is equal to $$\lim_{y\to 1} \frac{y^{m-1}+y^{m-2}+\cdots +1}{y^{n-1}+y^{n-2}+\cdots +1},$$ which is $\frac{m}{n}$.


2

How about this one?: $$f(n)=2\left\lfloor\frac n4\right\rfloor+1$$


2

Following user164587's comment: \begin{align} \lim_{x\to0^-}\frac1x&=-\infty\\ \lim_{x\to0^-}e^{1/x}&=0\\ \lim_{x\to0^-}2+e^{1/x}&=2\\ \lim_{x\to0^-}\frac1{2+e^{1/x}}&=\frac12\\ \lim_{x\to0^-}\arcsin\left(\frac1{2+e^{1/x}}\right)&=\arcsin\frac12\\ &=\frac\pi6 \end{align} Thus, we have ...


2

Edited: As pointed out in comments, previous proof was all wrong, should now be fixed! Let $x_0$ be the starting point, $x_1=f(x_0)$, $x_2=f(x_1)$, etc. We will show that if $x_n\in P_1$, there will be some later $x_{n+k}\in P_0$. This means there can't be an infinite tail of 1's, since every 1 is eventually followed by a 0. If $x_n\in P_1$, write ...


2

For instance, $f:[-1,1] \rightarrow \mathbb R$ defined as $f(x)=1/x$ if $x\neq 0$ and $f(0)=0$ is defined on a compact domain $[-1,1]$ but it is not bounded. Recall the Weierstrass theorem: "Every continuous function on a compact domain has at least one maximum and one minimum" So negating the above statement we obtain that: "No maximum or minimum and ...


1

It is simply the set of all integers ending with the digit $0$, $2$, $4$, $6$ or $8$. This is precisely the set of all even integers, or $\{2k:k\in\mathbb{Z}\}$.


1

notationwise: if $n \in \{0,...,9\}$ then $f^{-1}(n)=\{\pm(10^k+n)\}_{k \in \mathbb{N}}$ so (as explained by previous comments) $$ f^{-1}(\{0,2,4,6,8\}) = 2\mathbb{Z} $$


1

Take $$ f(x)=\mathrm{e}^x+x. $$ Then $$ f(0)=1>0 \quad\text{and}\quad f(-1)=-1+\frac{1}{\mathrm{e}}<0. $$ Hence, Intermediate Value Theorem provides that there exists a $x\in(-1,0)$, where $f$ vanishes.



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