I have tried to prove the following result, but looking around the internet and a few books, it always seems to be done differently - and I understand the variations I've seen. I would just like to know whether my argument below is logically correct or whether I have missed something out. It's been years since my Analysis modules at university and I'm afraid I'm a bit rusty at something I wasn't particularly great at at the time!  Thanks,  Andy.  Proposition (The Reciprocal Rule for Sequences)  If $(b_n)\rightarrow b$ and $b\neq 0$, then $\left(\frac{1}{b_n}\right)\rightarrow\frac{1}{b}$.  Proof.$\quad$ Suppose $b\neq 0$. Since $(b_n)\rightarrow b$, $(bb_n)\rightarrow b^2$. Because $\frac{b^2}{2}&gt;0$ we can find a natural number $K_1$ such that $|bb_n - b^2|&lt;\frac{b^2}{2}$ for all $n&gt;K_1$. This is equivalent to     \begin{equation*}   b^2-\frac{b^2}{2} &lt; bb_n &lt; b^2+\frac{b^2}{2},  \end{equation*}     in particular $bb_n&gt;\frac{b^2}{2}$, for all $n&gt;K_1$. Next, we see that     \begin{align*}   \left| \frac{1}{b_n}-\frac{1}{b} \right| &amp;= \left| \frac{b-b_n}{bb_n} \right|\\   &amp;&lt; \left| \frac{b-b_n}{\frac{b^2}{2}} \right|\\   &amp;= \left|\frac{2}{b^2}\right||b-b_n|\\   &amp;= \frac{2}{b^2}|b-b_n|\\  \end{align*}     for all $n&gt;K_1$. Let $\varepsilon&gt;0$. Next, since $(b_n)\rightarrow b$, $(b_n-b)$ is a null sequence. So we can find a natural number $K_2$ such that $|b_n-b|&lt;\frac{\varepsilon b^2}{2}$. Set $K=\max\{K_1,K_2\}$. Then     \begin{equation*}   \left| \frac{1}{b_n}-\frac{1}{b} \right| &lt; \frac{2}{b^2}|b-b_n| &lt; \frac{2}{b^2}\frac{\varepsilon b^2}{2} = \varepsilon  \end{equation*}     for all $n&gt;K$. Therefore $\left(\frac{1}{b_n}-\frac{1}{b}\right)\rightarrow 0$ and, equivalently, $\left(\frac{1}{b_n}\right)\rightarrow\frac{1}{b}.$       QED
While I was reading the book Exploratory examples for real analysis, I came across this:     to show that : $$\lim_{x\rightarrow x_o}\frac{1}{h(x)} = \frac{1}{\lim_{x\rightarrow x_o}{h(x)}} = \frac{1}{M}$$      Discussion:Let $\epsilon \gt 0$.we want to find $\delta \gt 0$ such that       if  $0 \lt |x-x_0| \lt \delta $,then $\Big|\frac{1}{h(x)}-\frac{1}{M}\Big|\lt \epsilon$.      In order to get expression involving $|h(x)-M|$,we first find the common denominator:        $\Big|\frac{1}{h(x)}-\frac{1}{M}\Big|=\Big|\frac{M-h(x)}{M~h(x)}\Big|$      $~=\Big|\frac{h(x)-M}{|M|~|h(x)|}\Big|$.      As we have to deal with variable term $|h(x)|$ in denominator.We can construct bound .Let $\epsilon =\frac{|M|}{2}$ ,since lim$_{x\to x_o}h(x)=M$,there exists $\delta_1\gt 0~~$ s.t.       if $0 \lt |x-x_0| \lt \delta_1 $,then $|h(x)-{M}|\lt \frac{|M|}{2}$.      As you study steps below,identify that step where $h(x)\neq 0$ for all $0 \lt |x-x_0| \lt \delta_1 $ plays a critical role:      $|M|=|M-h(x)+h(x)|$      $~~~~~~\leq |M-h(x)|+|h(x)|$      $~~~~~~= |h(x)-M|+|h(x)|$      $~~~~~~\lt \frac{M}{2}+|h(x)| \implies$      $\frac{|M|}{2} \lt |h(x)| \implies$      $\frac{1}{h(x)}\lt \frac{2}{|M|}$,where $0 \lt |x-x_0| \lt \delta_1 $.      Now that we've bound for $1/h(x)$,we can apply the definition to term $|h(x)-M|$.      Since     lim$_{x\rightarrow x_o}h(x)=M$,we can find $\delta_2 \gt 0$ such that,       $~~~~~~~~~~~~~~~~~~~~~~~~~~~~$if  $0 \lt |x-x_0| \lt \delta_2$, then $|h(x)-M|\lt \frac{|M|^2 \epsilon }{2}$   The thing I can't understand is for what purpose did the author work with such complicated expression involving $\epsilon $ in the end of the discussion?
How to install the previous version of a .deb package and pin it? I am trying to install the latest version of subversion on Sid, and because it has a bug I receive a warning and I abort the installation. How do I get to locate the previous version version, install it and pin until the bug is resolved?  root@server01:~# apt-get install subversion   Reading package lists... Done                                                                                                                                                                                  Building dependency tree                                                                                                                                                                                       Reading state information... Done                                                                                                                                                                              Suggested packages:                                                                                                                                                                                              db5.3-util subversion-tools                                                                                                                                                                                  The following NEW packages will be installed:                                                                                                                                                                    subversion                                                                                                                                                                                                   0 upgraded, 1 newly installed, 0 to remove and 205 not upgraded.                                                                                                                                               Need to get 0 B/981 kB of archives.                                                                                                                                                                            After this operation, 4,844 kB of additional disk space will be used.                                                                                                                                          Retrieving bug reports... Done                                                                                                                                                                                 Parsing Found/Fixed information... Done                                                                                                                                                                        critical bugs of subversion (-&gt; 1.9.2-2) &lt;Outstanding&gt;                                                                                                                                                          b1 - #803725 - subversion: dump-load of a repository modifies verbose log output: M line lost serious bugs of subversion (-&gt; 1.9.2-2) &lt;Outstanding&gt;  b2 - #803589 - FTBFS with ruby2.2 (only) Summary:  subversion(2 bugs) Are you sure you want to install/upgrade the above packages? [Y/n/?/...] n ********************************************************************** ****** Exiting with an error in order to stop the installation. ****** ********************************************************************** E: Sub-process /usr/sbin/apt-listbugs apt returned an error code (10) E: Failure running script /usr/sbin/apt-listbugs apt Table cells larger than they are meant to be I've created a map system for a game that runs on the principle of drawing the picture of the map from tiles. There are many reasons for this which I won't go into here but if you really want to know then I'm sure you can find out how to contact me ;)  I have  so you can see exactly where the problem lies and the source. The issue is the line between the top 2 tiles and the bottom 2 tiles, I can't figure out why it's gone like this and any help would be appreciated.  In the source is a marker called "stackoverflow", if you search for "stackoverflow" when viewing source then it should take you to the table in question.  I have also uploaded an . Proving an infinite number of primes of the form 6n+1 Unicode identifiers in Python? Are Newton's "laws" of motion laws or definitions of force and mass? When the phrase is used as an object, why so many native speakers are saying "you and I" instead of "you and me"? I'm not a native speaker but I thought "you and me" is correct. Not sure if this falls into the same category, but "Just between you and me" sounds more natural than "Just between you and I". Leave out a term when summing Z-axis issue with the floating review info panel, permits various elements to flow over it I was having a look at an old question of mine when I browsed the  from the timeline. I scrolled downwards and realized the Review completed panel comes along for the ride. While that might be , tags flowing over top of it as pictured below probably isn't.  Here's a .gif of the Question and Duplicate tabs flowing over it as well. Note that the timeline button also flows over. If vote buttons are present, they will also flow over it.  Reproduced on Chrome and Microsoft Edge. Update Per  by Tiago Martins Peres, this issue also happens in other Review types, namely Late Answers, First Posts and Low Quality Posts. In addition to that, it is causing the action buttons on reviews to become unclickable if you scroll low enough.  As a temporary workaround, you can use 's , which  this issue. How do I find the location of Python module sources? How do I learn where the source file for a given Python module is installed? Is the method different on Windows than on Linux?  I'm trying to look for the source of the datetime module in particular, but I'm interested in a more general answer as well. How does one actually show from associativity that one can drop parentheses? The force of gravity is constantly being applied to an orbiting object.  And therefore the object is constantly accelerating.  Why doesn't gravity eventually "win" over the object's momentum, like a force such as friction eventually slows down a car that runs out of gas?  I understand (I think) how relativity explains it, but how does Newtonian mechanics explain it? Irreducible Components of the Prime Spectrum of a Quotient Ring and Primary Decomposition How do I save the output of a command to a file?  Is there a way without using any software? I would like to know how. Why are all of Mr. Shirley's gifts the same? References in Chapter Titles I used to say "sharp sign" to refer to the # sign.   Today a friend told me that the correct term is number sign or hash sign or even just hash.  What is the difference between these options and what's the correct usage of the names for this sign? Relevance of cycle detection in Pollard's rho integer factorization algorithm I've recently learned about Pollard's rho factorization algorithm, and the way I understand it, it works by using a pseudorandom sequence of integers $x_{k}$ ranging from $0$ to $n-1$, where $n$ is the number to be factored, to find repetitions in an unknown sequence $y_{k} = x_{k} \bmod p$, where $p$ is an unknown, non-trivial factor of $n$. Such a repetition $y_{i}=y_{j}$ occurs when $x_{i}\equiv x_{j} \bmod p$, which can be detected by checking whether $\gcd(x_{i}-x_{j}, n)\gt 1$.  From what I could find on the Internet, Floyd's cycle detection algorithm or a variant thereof is used to find such repetitions. And the reason for this is what I don't understand.   First, Floyd's algorithm requires that each element in the sequence depends solely on the element before it, which, even if it is the case for $x_{k}$, is not necessarily the case for $y_{k}=x_{k}\bmod p$. For example, assuming we want to factorize $6$, let's take the sequence $5, 3, 2, 4, 1, 0, 5, 3, 2, 4, 1, 0,\ldots$. The cycle period is $5, 3, 2, 4, 1, 0$, where every element is contained only once. However, if we take this sequence $\bmod 2$, we get $1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0,\ldots$. Floyd's algorithm cannot be applied here to find the cycle period $1, 1, 0, 0, 1, 0$. Besides, we are not even interested in a cycle, only in repetitions. Whether we find the congruence $\bmod 2$ between $5$ and $3$, $5$ and $1$, or $4$ and $2$ in the first sequence is irrelevant.   Please correct me if my understanding of Pollard's rho algorithm as described above is wrong. But if it is not, it seems to me that one could just as well try out some random numbers between $2$ and $n-1$ and see if they share a divisor with $n$ other than $1$. So what is the benefit of using a cycle detection method like Floyd's algorithm?  I also found , but it is not answered, and the comment on the question does not really answer the question either. Can't Use /Summon Command How to restore original size and unboot my bootable USB pendrive 10% rule for sample sizes In an introductory stats book by Nicole Radziwell "Statistics the easy way with R" , an assumption used  for nearly every statistical test (e.g.t-tets, anova, etc) is that the sample size should not be more than 10% of the population size (where the population size in known).   I havent seen it mentioned in other introductory stats books. My question is: how important is theis assumotion and what is the reason for it?  The author didn't say where this comes from (at least as far as I could see). A bit of googling suggests that it refers to making a simple random sample without replacement, if the sample size is above 10% then the observations wont be (approximately) independent, and indepedence of observations is an assumption stated in most textbooks.   In the case where you have a small population (e.g. &lt;50), should you give up on this assumption because it results in sample sizes too small to get decent estimates of variability and to check other assumptions such as normality.    Thank you for any insights.