$f(x,y) := \begin{cases} \frac{xy}{x^2+y^2}, &amp;\mbox{if } (x,y) \neq (0,0)\\  0, &amp; \mbox{if } (x,y)=(0,0) \end{cases}$  I need to show that all partial derivatives in $(0,0)^T$ exist which I already did, but how can I show that $f$ is not differentiable in $(0,0)^T$?
Let $F:\mathbb{R^2}\rightarrow \mathbb{R}$ be a function, such as : $$F(x,y)=\begin{cases}        0 &amp; (x,y)=(0,0) \\       \frac{xy}{x^2+y^2} &amp; (x,y)\neq (0,0)    \end{cases}$$      Is $F$ differentiable at $(0,0)$?   My attempt :  Consider the limit, $\lim_{(x,y)\rightarrow (0,0)} F(x,y)$ along the line $y=x$. The limit equals to $\frac{1}{2}$. However, $F(0,0)=0$. Hence $F$ is not continuous at $(0,0)$ which implies that $F$ isn't differentiable as well.  Is this approach okay ?
Does Stack Exchange use caching and if so, how? Non-hidden directory doesn't show in Explorer, but the program that made it can access it I have multiple Ubuntu machines at home and a pretty slow internet connection, and sometimes multiple machines need to be updated at once (especially during new Ubuntu releases.)  Is there a way where only one of my machines needs to download the packages, and the other machines can use the first machine to get the debs?  Does it involve setting up my own local mirror?  Or a proxy server?  Or can it be made simpler? Why exactly do sometimes universal covers, and sometimes central extensions feature in the application of a symmetry group to quantum physics? There seem to be two different things one must consider when representing a symmetry group in quantum mechanics:   The universal cover: For instance, when representing the rotation group $\mathrm{SO}(3)$, it turns out that one must allow also $\mathrm{SU}(2)$ representations, since the negative sign a "$2\pi$ rotation" induces in $\mathrm{SU}(2)$ is an overall phase that doesn't change the physics. Equivalently, all representations of the Lie algebra are what we seek. ($\mathfrak{so}(3) = \mathfrak{su}(2)$, but although every representation of the algebra is one of the universal cover, not every representation of the algebra is one of $\mathrm{SO}(3)$.) Central extensions: In conformal field theory, one has classically the  of infinitesimal conformal transformations. From the universal cover treatment one is used to in most other cases, one would expect nothing changes in the quantum case, since we are already seeking only representation of an algebra. Nevertheless, in the quantization process, a  appears, which is often interpreted to arise as an "ordering constant" for the now no longer commuting fields, and we have to consider the  instead.   The question is: What is going on here? Is there a way to explain both the appearence of universal covers and central extensions in a unified way? Max Planck, arguably the leading theoretical Physicist of his day, apparently used of experimental data to formulate his equation for , which was a major step in the birth of quantum mechanics. For this seminal event of the 20th century, I would expect a detailed explanation of how Planck arrived at his conclusion that energy MUST be quantized. I've even read his Nobel lecture but was non the wiser.  Can some kind person please explain it, or give me even just a hint? Is it possible to use the startup.blend UI layout in older files? Is there a way to open files without changing the UI layout?  Sometimes in older files the layout will be in the old 2.4x style, which can be annoying.  One way would seem to be appending everything in, but that gets tedious with having to do it over again and again for objects, materials, textures, etc. How do World of Warcraft referrals and the free month work? I sent the referral link to my boss and he bought the game and applied his CD key.  He now has a paid account linked to mine:    Yet when I view my World of Warcraft account, it seems they didn't give me the free month. It appears I only have the free month that came with my Battlechest purchase.  I just created and bought my account this morning. Would this affect the free referral month in any way?    We haven't seen each other or partied online in the game yet, do we need to do that first in order for me to receive the month? I am using a premium legit version of Minecraft that I purchased. I can log into minecraft.net as well as mojang.com but I am unable to login to Minecraft the game through the minecraft.exe or the Technic launcher. The Minecraft servers are up and running fine. My internet connection is also fine. I double checked my credentials and even tried changing my password. I tried using both my username and my email. I tried all of this on two separate accounts that I own. I tried deleting my login, launcher, launcher.pack.lzma, and launcher_profiles.json and then reopening Minecraft. I tried re-downloading minecraft.exe.   Nothing has worked and I seem to have run out of options. I sent a help request to Mojang but in the meantime I was hoping that maybe you guys could offer some advice. $$\tan(\theta_1+\theta_2+\cdots+\theta_n)=\frac{S_1-S_3+S_5+\cdots}{1-S_2+S_4+\cdots}$$  where  $S_i$ denotes the sum of product of tangent of angles taken $i$ at a time.     For example,$$\tan(\theta_1+\theta_2)=\frac{S_1}{1-S_2}=\frac{\tan\theta_1+\tan\theta_2} {1-\tan\theta_1\tan\theta_2}$$   (This formula is given  in my textbook with no derivation or background)  How to derive this? How do I delete a Git branch locally and remotely? Calling the base constructor in C# I have been wondering why anyone ever thought that we could find an expansion for eigenstates of some perturbed Hamiltonian in terms fo those for the basis Hamiltonian. My lecturer insisted that this was because the eigenstates of the Hamiltonian are complete', however from my (limited) understanding of the mathematical basis of QM:   When you are given a Hamiltonian, it comes with the Hilbert space of functions of which it acts (subject to some conditions we impose, such as the Lebesgue square-integrability of functions) When you are given another Hamiltonian, there is no guarantee that it will act on the same Hilbert space Therefore, that the zeroth order Hamiltonian has eigenstates forming a complete basis for the space on which ti acts does not guarantee that these same eignestates form a complete basis for the Hilbert space on which the purturbed Hamiltonian acts.   As a simple example. I can imagine adding a small perturbing magnetic field to a Hamiltonian in which there initially is none, such that we lose the degeneracy in the eigenstates accounting for different spins of a particle. One could say that we have the same Hilbert space before and after but we lose degeneracy; or I would have been tempted to say that our initial Hilbert space did not have these extra dimensions corresponding to the spin states. So rather than the eigenstates of the basis Hamiltonian also forming a complete basis for the purtubed Hamiltonian, but having eigenvalue degeneracy which is lifted on the purtubration, I would have said that our Hilbert space has changed. After all, if we take the former view (i.e. a basis accouting for spin but with degeneracy) then it seems like for any problem we would need to know and account for EVERY possible system property that can be discerned. This doesn't seem reasonable to me....  NOTE:  1) My question was asked here previously:   although there is only a discussion and no conclusive answer given. I see that the distinction is important in relativistic QM, and the case of adding a potential onto a free particle Hamiltonian seems to be a special one. Although I am still not sure what the answer is for general purturbations.  2) Similar question here:   but it seems to completely avoid answering the question. What is mirror lockup and what is its primary function? My SLR is equipped with the mirror lockup function. What does it do, and when would I employ such a feature? Is there an idiom that means "revealing a secret unintentionally"? I am thinking there could be an idiom that states that you or someone revealed a secret unintentionally? Can you think of an idiom like that? I am looking for an idiom I can use to write a blog post about the recent incident where Secretary Pompeo unintentionally revealed the dirty secrets behind the CIA. Can I buy a plane ticket from A-B-C but get on the plane at B? The "proper" way to test if a service is running in a script My problem:  I'm writing a bash script and in it I'd like to check if a given service is running.  I know how to do this manually, with $ service [service_name] status.  But (especially since the move to systemd) that prints a whole bunch of text that's a little messy to parse. I assumed there's a command made for scripts with simple output or a return value I can check.  But Googling around only yields a ton of "Oh, just ps aux | grep -v grep | grep [service_name]" results. That can't be the best practice, is it? What if another instance of that command is running, but not one started by the SysV init script?  Or should I just shut up and get my hands dirty with a little pgrep? difference between over and for I have a simple question.     over centuries / for centuries   what is the difference?  Thank you. In Drush 8, my site aliases are loaded correctly from ~/.drush/mysite.aliases.drushrc.php.  E.g.  $aliases['dev'] = array(   'root' =&gt; '/var/www/html/mysite.dev/docroot',   'uri' =&gt; 'mysite.prod.acquia-sites.com',   'remote-host' =&gt; 'mysite.ssh.acquia-sites.com',   'remote-user' =&gt; 'mysite.dev',   'path-aliases' =&gt; array(     '%drush-script' =&gt; 'drush' . $drush_major_version,   ), );   However these aliases aren't loaded in Drush 9.  $ drush site-alias {  }   How I should define site aliases in Drush 9? Is there any migration process? 0 ram free on ICS This is a revisit of  that seems to have grown stale without any solid answer. Just recently I was baffled by 's ability to :  In my experience, certain review queues seem to be more inclined to &quot;single-user reviews&quot; (like First Posts and/or Late Answers), while others are more geared to &quot;multi-user reviews&quot; (like Close Votes and/or Suggested Edits). This does not seem to be the case here. The &quot;requirement&quot; for such late, duplicate reviews are questionable and               (         ). Even being able to  (I highly doubt this was an instance of keeping the review window open for more than an hour and only reviewing it post mortem - first post reviews are dealt with far more quickly than most other review queues):   While these examples do not speak to the , it sure seems biased. Something is amiss... Over at  we have a far smaller community, so these situations are far more noticeable.  Edit: The following are all candidates for the above discussion: