Question: Find  the value $$A=\lim_{x \rightarrow 0 }(\frac{(1+x)^{1/x} - e  + \frac{ex}{2}}{x^2})$$  I found the value $$A=\lim_{x \rightarrow 0 }(\frac{(1+x)^{1/x} - e  + \frac{ex}{2}}{x^2})=\lim_{x \rightarrow 0 }( \frac{e- e + ex/2}{x^2}) =0  $$ Is this correct ?
$$\lim_{x\rightarrow 0}\frac{(1+x)^{1/x}-e+\frac{ex}{2}}{x^2}=\,?$$   by directly substituting $x=0$ i got $\infty$ by using L-H's rule, i got $-1/8$   the given options are  $a)\frac{24e}{11}$    $b)\frac{11e}{24}$    $c)\frac{e}{11}$    $d)\frac{e}{24}$    may be the question would be  $$\lim_{x\rightarrow 0}\frac{(1+x)^{1/x}-e+\frac{ex}{2}}{x^2}=\,?$$
Does $\lim_{h\rightarrow 0}\ [f(x+h)-f(x-h)]=0$ imply that $f$ is continuous? Surjectivity of Composition of Surjective Functions Suppose we have two functions, $f:X\rightarrow Y$ and $g:Y\rightarrow Z$. If both of these functions are onto, how can we show that $g\circ f:X\rightarrow Z$ is also onto? pgAdmin to display multiple result sets Show that if $ab$ has finite order $n$, then $ba$ also has order $n$. - Fraleigh p. 47 6.46. Attempted proof based on  and .  Given: $a,b$ elements of $G$, and $ab$ has finite order $n$.   Hence $\color{magenta}{|ab| = n} \iff (ab)^n = e$.   I must prove that $n$ is the smallest positive integer        such that $(ba)^n = e.$ Notice:  $\begin{align}(\color{darkorange}{a}\color{darkcyan}{b})^n &amp;= e \implies \color{darkorange}{a}\color{darkcyan}{b}(ab)^{n-1} = e \implies  \color{darkcyan}{b}(ab)^{n-1}\color{darkorange}{a} = e \implies (ba)^n = e \end{align} $ $\implies |ba| \le n$.  Now I must prove that there's no positive integer $m$ such that $m&lt;n$ and $(ba)^m = e.$ For a proof by contradiction, suppose $|ba|&lt;\color{magenta}{n}$. Then we could apply the same reasoning to find that $ |ab| ≤ |ba| &lt; \color{magenta}{|ab|}$,  which is absurd. So $|ba| = |ab| = n. \quad \blacksquare$  Questions   Whence does $ |ab| ≤ |ba| $ arise, in the second last line?   What's the intuition of the claim proved?    How do you prognosticate that $|ba|&lt;\color{magenta}{n}$ is impossible, and that you should prove it by contradiction? Compiling C code in the terminal According to two questions  and  asked on this site earlier there exists a nice relation:  $$\frac1{n!} \left(\int_{0}^t\mathrm dt \; f(t)\right)^n = \int_{0}^t\mathrm dt_1 \int_{0}^{t_1}\mathrm dt_2 \cdots \int_{0}^{t_{n-1}}\mathrm dt_n\; f(t_1)\,f(t_2) \ldots f(t_n). $$  Unfrotunatly I am more interested in integrals of the form  $$\int_{0}^t\mathrm dt_1 \int_{0}^{t_1}\mathrm dt_2 \cdots \int_{0}^{t_{n-1}}\mathrm dt_n\; f(t_n). $$  Can it be somehow brought to an unary integral?  All I have figured out by now is that  $$\int_{0}^t\mathrm dt_1 \int_{0}^{t_1}\mathrm dt_2 \cdots \int_{0}^{t_{n-1}}\mathrm dt_n\; t_n^m = \frac{m!}{(m+n)!} t^{m+n} .$$ My girlfriend forgot the password for her MacBook (don't ask), so I reset it using the Utilities.  But as you may be aware, this doesn't reset the Keychain password. Her MacBook is now virtually unusable due to Keychain demands continuously provoking a request for the Keychain password.  There are lots of other people asking this around the web and they're usually pointed towards this , which doesn't apply in our situation because the support advice involves entering the old password.  We don't know the old password.  Is it possible to reset Keychain without knowing the old password? We're not bothered about what's in there and what will be lost, we just want the continuous password prompts to go away. Nice proofs of $\zeta(4) = \frac{\pi^4}{90}$? I know some nice ways to prove that $\zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \pi^2/6$.  For example, see  or the answers to the question "?"  Are there any nice ways to prove that $$\zeta(4) = \sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}?$$  I already know some proofs that give all values of $\zeta(n)$ for positive even integers $n$ (like #7 on Robin Chapman's list or Qiaochu Yuan's answer in the linked question).  I'm not so much interested in those kinds of proofs as I am those that are specifically for $\zeta(4)$.  I would be particularly interested in a proof that isn't an adaption of one that $\zeta(2) = \pi^2/6$. Is it just for me, or has the privilege to review new posts and late answers changed for everyone?  I recently started being more active in reviewing, but the rep requirement just shot up and I can not review anymore? The rep requirement has shot up to 500. Is this just for me, as a 'process", or has it been changed site wide? Fourier series identity I need to prove that $\dfrac{a \sin(bx)}{1 - 2a \cos(bx) + a^2} = \sum_{n=1}^\infty a^n \sin(nbx)$ where $|a| &lt; 1$.  It seems that this can be proved by using Euler's formula identities for $\cos(bx)$ and $\sin(bx)$ and substituting $z = e^{ibx}$. From that, I get $$ \dfrac{a \sin(bx)}{1 - 2a \cos(bx) + a^2} = \dfrac{a (z - 1/z)}{2i(1 - a(z + 1/z) + a^2)}$$ $$= \dfrac{a (z^2 -1)}{2i(z - a(z^2 +1) + za^2)}$$ $$= \dfrac{a (z^2 -1)}{-2ai(z^2 - z(1-a^2)/a + 1)}$$  I don't understand what I am supposed to do next. Floor function properties: $[2x] = [x] + [ x + \frac12 ]$ and $[nx] = \sum_{k = 0}^{n - 1} [ x + \frac{k}{n} ] $ I'm doing some exercises on Apostol's calculus, on the floor function. Now, he doesn't give an explicit definition of $[x]$, so I'm going with this one:     DEFINITION Given  $x\in \Bbb R$, the integer part of $x$ is the unique $z\in \Bbb Z$ such that $$z\leq x &lt; z+1$$ and we denote it by $[x]$.   Now he asks to prove some basic things about it, such as: if $n\in \Bbb Z$, then $[x+n]=[x]+n$  So I proved it like this: Let $z=[x+n]$ and $z'=[x]$. Then we have that  $$z\leq x+n&lt;z+1$$  $$z'\leq x&lt;z'+1$$  Then $$z'+n\leq x+n&lt;z'+n+1$$  But since $z'$ is an integer, so is $z'+n$. Since $z$ is unique, it must be that $z'+n=z$.  However, this doesn't seem to get me anywhere to prove that  $$\left[ {2x} \right] = \left[ x \right] + \left[ {x + \frac{1}{2}} \right]$$  in and in general that   $$\left[ {nx} \right] = \sum\limits_{k = 0}^{n - 1} {\left[ {x + \frac{k}{n}} \right]} $$  Obviously one could do an informal proof thinking about "the carries", but that's not the idea, let alone how tedious it would be. Maybe there is some easier or clearer characterization of $[x]$ in terms of $x$ to work this out.  Another property is $$[-x]=\begin{cases}-[x]\text{ ; if }x\in \Bbb Z \cr-[x]-1 \text{ ; otherwise}\end{cases}$$  I argue: if $x\in\Bbb Z$, it is clear $[x]=x$. Then $-[x]=-x$, and $-[x]\in \Bbb Z$ so $[-[x]]=-[x]=[-x]$. For the other, I guess one could say:  $$n \leqslant x &lt; n + 1 \Rightarrow  - n - 1 &lt; x \leqslant -n$$  and since $x$ is not an integer, this should be the same as $$ - n - 1 \leqslant -x &lt; -n$$  $$ - n - 1 \leqslant -x &lt; (-n-1)+1$$  So $[-x]=-[x]-1$ Why is the picture not inserted? When I pdflatex the text below it doesn't insert the picture, but writes the name of the file instead (pic).  And the geometry line gives the error  ! LaTeX Error: Missing \begin{document}.   The picture I am using is at  and the floatrow package .  Question  Can anyone see what is wrong?  Text  \documentclass[a4paper]{article} \pdfpagewidth=\paperwidth \pdfpageheight=\paperheight \usepackage{aeguill} \pdfcompresslevel=9 \usepackage[pdftex,colorlinks]{hyperref}  \usepackage{a4} \usepackage{fixltx2e} \usepackage[danish]{babel} \usepackage[latin1]{inputenc} \usepackage[T1]{fontenc}  \usepackage[draft]{graphicx} \usepackage{floatrow}  \usepackage{geometry}  \geometry{margin=10pt,paperwidth=\textwidth+20pt,paperheight=\textheight+20pt}  \begin{document}  The first paragraph.  \begin{figure}   \includegraphics{pic} \end{figure}  And the second.  \end{document} Not sure this is something which people will know on here are not.    We currently have a BI setup where we have Integration services on one SQL instance and the data warehouse database on a separate SQL Instance.    My question is, as no users physically connect to the Integration services SQL instance and it is purely used to execute package from the integrations service catalog via SQL agent, do we need to License every user i.e. Cal's for that SQL instance.  Basically i'm wanting to know if I could utilize SQL 2016 for an integration services instance and use features like Json manipulation etc and then load this data into the 2014 version instance...but what level of licensing will I need? How to determine equality for two JavaScript objects? I posted a question and a couple of respondents asked for the source code. I could only post portions of the code.  If attachments were allowed I could have uploaded a .zip file with the whole application.  Will this be a feature in SO? Run bash script at login stored in the home folder? When I try to load a LaunchAgent plist from launchctl I can't find out how to run a script in the home directory.  My code is:  &lt;?xml version="1.0" encoding="UTF-8"?&gt; &lt;!DOCTYPE plist PUBLIC "-//Apple//DTD PLIST 1.0//EN" "http://www.apple.com/DTDs/PropertyList-1.0.dtd"&gt; &lt;plist version="1.0"&gt; &lt;dict&gt;     &lt;key&gt;ProgramArguments&lt;/key&gt;     &lt;array&gt;         &lt;string&gt;bash&lt;/string&gt;         &lt;string&gt;~/script.sh&lt;/string&gt;     &lt;/array&gt;     &lt;key&gt;RunAtLoad&lt;/key&gt;     &lt;true/&gt;     &lt;key&gt;Label&lt;/key&gt;     &lt;string&gt;com.tyilo.test&lt;/string&gt; &lt;/dict&gt; &lt;/plist&gt;   I have tried both with and without bash and also replacing ~ with $HOME. I have also tried using bash -c without it working.  The error code is:   `com.tyilo.test: bash: ~/script.sh: No such file or directory` Increase iPhone call history How to shrink Windows 7 boot partition with unmovable files I have just bought an HP laptop with Windows&nbsp;7 (64 bit). It has a 500&nbsp;GB HDD with three partitions: a small hidden system partition, a 12&nbsp;GiB HP recovery partition, and a 450&nbsp;GiB C: boot partition.  I would like to split this large C: partition into two partitions, leaving only 100&nbsp;GiB for the system, and giving the rest to a new data partition.  Although the Windows built-in Disk Management utility has an option to shrink the bootable partition, it only allows me to shrink it roughly by half, even though only 20&nbsp;GiB on the partition is used. As far as I understand, system unmovable files lie in the middle of the partition, preventing Disk Management utility to do what I want.    And since new HP laptops don't come with OS installation disks (they only allow you to create recovery disks yourself), I can't just repartition HDD and then reinstall OS.  So, is there a way to shrink the C: bootable partition and preserve Windows&nbsp;7 working?  P.S.: I have tried to use the third-party  utility, and after shrinking the partition, Windows&nbsp;7 stopped booting, with a . System recovery didn't work, and I had to do factory recover. Since this is a long process, I would like to avoid doing it again :) So, please, suggest only proven solutions. Should "each" be followed by a singular or plural possessive? If a possessive noun, which is plural, is preceded by "each", then should it use the singular or plural possessive form?  For example, which of the following is correct?    spend time in each other's presence spend time in each others' presence How to handle repeated references in footnotes?