I am trying to solve this limit without using Taylor expansions to explain it to a secondary school student   $$\lim_{x \to 0} \frac{10^x - 2^x - 5^x + 1}{x \cdot \tan x}.$$
$$\lim_{x\to 0}\frac{10^x - 2^x - 5 ^ x + 1 } {x\tan x} $$ This is an indeterminate limit.  I want help in solving this problem. Thanks in advance
The Windows 8 version of Google Chrome seems to have been updated to support the UI formerly known as Metro.  However, this seems to be the only option now, which means that I can no longer use Chrome on my netbook.  How do I get Google chrome to work in classic desktop mode on Windows 8?  Thanks I was wondering why a laser beam diverges. If all the photons are in the same direction, I would imagine that it would stay that way over a long distance. I am aware that a perfectly collimated beam with no divergence cannot be created due to diffraction, but I am looking for an explanation based on  photons rather than wave physics. Can't login with Google using Internet Accounts How many throws of a die until every possible result is obtained? A die is thrown until every possible result (i.e., every integer from 1 to 6) is obtained. Find the expected value of the number of throws.   How do I do that? I understand that probability for the single result is $\{1, 5/6, \ldots , 1/6\}$, but what about the expected value? Creating Workflow -> Error When I try to create a new WF in my Sharepoint Designer I get following error:     The list of workflow actions on the server references an assembly that   does not exist. Some actions will not be avaliable. The assembly   strong name is "Microsoft.SharePoint.WorkFlowActions,   Version=14.0.0.0,Cultureneutral,PublickeyToke=   Does any on know how to fix this problem? still frame camera solve I've found a number of how-tos for solving video but I'm just looking to put an element in a photo and can't seem to find a how-to for that.  Is it possible in blender to do a camera solve with a still image?  Thanks. Auto-inserting Stack Overflow affiliate into all Amazon links We have officially discontinued the Amazon book affiliate remnant ads, since despite our best efforts they don't perform well for our audience and cannot be made to perform well.  We've often wondered,   I'm thinking this isn't a problem; it has gotten lots of support historically. So we're going to go ahead and do it, so the company Amazon account is still getting some use, and we benefit a little.   To be clear: from this point on, all Amazon links posted on all sites will automatically be rewritten to add our affiliate link. We have a clever affiliate redirect that geolocates the IP address and redirects to the appropriate regional Amazon store (amazon.uk, amazon.ca, et cetera).  If you have any feedback on this, or suggestions, as they say &hellip; speak now or forever hold your peace! Suppose that $f:\Bbb R\to \Bbb R$ is continuous and that $f(x)\to 0$ as $x\to \pm\infty.$Prove that $f$ is uniformly continuous. Suppose that $f:\Bbb R\to \Bbb R$ is continuous and that $f(x)\to 0$ as $x\to \pm\infty$. Prove that $f$ is uniformly continuous.   Since $f(x)\to 0$ as $x\to \pm\infty$ we have $|f(x)|&lt;\epsilon $ whenever $x&lt;-K$ and $x&gt;K$ where $K&gt;0$.  Also $f$ is uniformly continuous on $[-K,K]$ as the domain is compact.  Hence $f$ is uniformly continuous in $(-\infty,-K)\cup [-K,K]\cup (K,\infty)$.  But how to conclude that $f$ is uniformly continuous on $\Bbb R$ from above as I may chose $x\in (-\infty,-K);y\in [-K,K]$ ,how to conclude that $|f(x)-f(y)|&lt;\epsilon $ from above for a given $\epsilon$?  Please suggest the edits required. A set with a finite integral of measure zero? Prove $(x_1+.....+x_n)^2 \leq n(x_1^2 + .....+x_n^2)$ Prove that  $(x_1+\dots+x_n)^2 \leq n(x_1^2 + \dots+x_n^2)$  for all positive integers n and all real numbers $x_1,....,x_n$  I am attempting a proof by induction but wasn't sure if i need the Cauchy-Schwarz Inequality or perhaps another way other than induction to prove this.  Proof   $n=1$ true  assume true for $n=k$  Now for $n= k+1$  $(x_1 + \dots +x_k + x_{k+1})^2 \leq \dots$ Number of equivalence relations on a set Solving of a seperable differential equation How do you solve this equation:  $$\frac{\mathrm dy}{\mathrm dx} + \frac{1}{xy} = y^3$$  I've tried to multiply everything by $y$ but I can not get all $x$'s ($\mathrm dx$) on one side and $y$'s ($\mathrm dy$) on the other. After taking a statistics course and then trying to help fellow students, I noticed one subject that inspires much head-desk banging is interpreting the results of statistical hypothesis tests.  It seems that students easily learn how to perform the calculations required by a given test but get hung up on interpreting the results.  Many computerized tools report test results in terms of "p values" or "t values".  How would you explain the following points to college students taking their first course in statistics:   What does a "p-value" mean in relation to the hypothesis being tested?  Are there cases when one should be looking for a high p-value or a low p-value? What is the relationship between a p-value and a t-value? Suddenly mysql/mariadb won't start Test if a package (or package option) is loaded [french]{babel} (aka frenchb) is incompatible with some other packages, such as tikz, since it tries to insert \thinspace before semi-colons.  In order to make these two play nice, you have to call \shorthandoff{;} in the document.  Now I have a package that uses tikz and I would like to locally disable the semi-colon for babel, instead of disabling it for the whole document. I could hack the .sty of the package to do that, knowing that I'm using frenchb, but I would rather make something clean and portable.  Is there a way in my package that I can test if frenchb is loaded and disable semi-colon management with babel locally, so the package doesn't have to be hacked?  Edit:  Martin's solution gives me something like this in general:  \RequirePackage{ifthen} \newcommand{\@babel@loaded}{0} \AtBeginDocument{\@ifpackageloaded{babel}                                                    {\renewcommand{\@babel@loaded}{1}}                                                         {\renewcommand{\@babel@loaded}{0}}                                                       } \newcommand{\somecommandusingsemicolon}{   \ifthenelse{\@babel@loaded=1}{\shorthandoff{;}}{}     % code with semi-colon here }   and in the specific case of tikz:  \AtBeginDocument{\@ifpackageloaded{babel}   {%     \g@addto@macro\tikz@installcommands{%       \shorthandoff{;}%     }     \g@addto@macro\tikz@uninstallcommands{%       \shorthandon{;}%     }   }   {\relax} } Algebraic proof of $\sum_{i=0}^k{{n \choose i}{m \choose {k-i}}}= {{m+n}\choose k}$ I can't figure out an algebraic proof for the following identity:  $$\sum_{i=0}^k{{n \choose i}{m \choose {k-i}}}= {{m+n}\choose k}$$  Combinatorical solution:  We can see that as choosing some from $n$ and the rest of $k$ from $m$, thus $k$ in total.  Or we could just choose $k$ from the union. I have a table with a varchar(max) column that stores text ~4-10KB in size. Data is just being added - never modified, it is a log. My problem is that a table grows too large in size - I would like to keep about 1000000 rows in this table.   I am limited to SQL Server Express and hit the limit on database size. Compression is not available in SQL Server Express and I afraid that Filestream may not be appropriate and the number of files would be too big.   What is the best solution in this case? The setuid permission bit tells Linux to run a program with the effective user id of the owner instead of the executor:  &gt; cat setuid-test.c  #include &lt;stdio.h&gt; #include &lt;unistd.h&gt;  int main(int argc, char** argv) {     printf("%d", geteuid());     return 0; }  &gt; gcc -o setuid-test setuid-test.c &gt; ./setuid-test  1000  &gt; sudo chown nobody ./setuid-test; sudo chmod +s ./setuid-test &gt; ./setuid-test  65534   However, this only applies to executables; shell scripts ignore the setuid bit:  &gt; cat setuid-test2  #!/bin/bash id -u  &gt; ./setuid-test2  1000  &gt; sudo chown nobody ./setuid-test2; sudo chmod +s ./setuid-test2 &gt; ./setuid-test2  1000     Wikipedia :     Due to the increased likelihood of security flaws, many operating systems ignore the setuid attribute when applied to executable shell scripts.   Assuming I'm willing to accept those risks, is there any way to tell Linux to treat the setuid bit the same on shell scripts as it does on executables?  If not, is there a common workaround for this problem? My current solution is to add a sudoers entry to allow ALL to run a given script as the user I want it run as, with NOPASSWD to avoid the password prompt. The main downsides to that is the need for a sudoers entry every time I want to do this, and the need for the caller to sudo some-script instead of just some-script Adjust space between entries in LoF and LoT Let $f\in C^{\infty}(\mathbb{R},\mathbb{R})$ such that $f(x)=0$ on $\mathbb{R}$\ $(-1,1)$. Prove that there exists $n\in\mathbb{N}$ such that $f^{(n)}$ has at least n+1 zeros on $(-1,1)$  My attempt :   Assume $f$ is strictly positive  (if not n=0 works). Let $Z(f^{(n)})$ the number of zeros of $f^{(n)}$ on $(-1,1)$. Then we have: $$ \forall n\in\mathbb{N}:\;f^{(n)}(-1)=f^{(n)}(1)=0 $$ so that iterating the Rolle theorem to the successive derivatives of f we have  $$  \;Z(f^{(k+1)}) \geq Z(f^{(k)}) +1 $$ thus, by induction : $$ \forall n\in\mathbb{N}: \;Z(f^{(n+1)}) \geq Z(f^{(n)}) +1 $$ Therefore, $$ Z(f^{(n)}) \geq n $$ How can I find an another zero ?   I'm still looking for an another proof which does not use the lemma ..,  Thank you in advance.