How would I show that $$f(x,y) = \frac{2xy}{x^2 + y^2}$$  is not differentiable at the origin?   Is it enough to show that as the function tends to the origin along the paths $y = x$ and $y=2x$ that we get different limits and hence the function is not continuous?
$$f(x,y) = \left\{\begin{array}{cc} \frac{xy}{x^2+y^2} &amp; (x,y)\neq(0,0) \\ f(x,y) = 0 &amp; (x,y)=(0,0) \end{array}\right.$$  In order to verify if this function is differentiable, I tried to prove it by the theorem that says that if $\frac{∂f}{∂x}$ and $\frac{∂f}{∂y}$ exist and are continuous at the point $(x_0,y_0)$, then the function is differentiable at this point. So I did:  $$\frac{\partial f}{\partial x}(0,0) = \lim_{h\to 0}\frac{f(0+h,0)-f(0,0)}{h} = 0$$ $$\frac{\partial f}{\partial y}(0,0) = \lim_{h\to 0}\frac{f(0,0+h)-f(0,0)}{h} = 0$$  so we have that the partial derivatives at point $(0,0)$ is $0$. Now, if we take the derivative at $(x,y)\neq (0,0)$ and then take the limit of it as $(x,y)\to(0,0)$, we can see if the derivatives are continuous or not. So here it is:  $$\frac{\partial f}{\partial x}(x,y) = \frac{y(y^2-x^2)}{(x^2+y^2)}$$  but   $$\lim_{(x,y)\to(0,0)} \frac{y(y^2-x^2)}{(x^2+y^2)} $$ does not exist (by wolfram alpha... but can anybody tell me an easy way to prove this limit does not exist? easier than taking the limit in different directions?), therefore the derivative is not continuous at $(0,0)$, so we can't say $f$ is differentiable at $(0,0)$, but for $(x,y)\neq (0,0)$ the function is continuous, as it is a quotient of continuous functions. So $f$ is at least differentiable at $(x,y)\neq (0,0)$.   Now, to verify differentiability at $(0,0)$ I think we must use the limit definition of differentiablity: A function is differentiable at $(0,0)$ iff: $$\lim_{(h,k)\to (0,0)} \frac{f(0+h,0+k)-f(0,0)-\frac{\partial f}{\partial x}(0,0)-\frac{\partial f}{\partial y}(0,0)}{\|(h,k)\|} = 0$$  Let's calculate this limit:  $$\lim_{(h,k)\to (0,0)} \frac{f(0+h,0+k)-f(0,0)-\frac{\partial f}{\partial x}(0,0)-\frac{\partial f}{\partial y}(0,0)}{\|(h,k)\|} = \\ \lim_{(h,k)\to (0,0)} \frac{\frac{hk}{h^2+k^2}}{\sqrt{h^2+k^2}} = \\ \lim_{(h,k)\to (0,0)} \frac{hk}{(h^2+k^2)\sqrt{h^2+k^2}}$$  which I think, it's a limit that does not exist, therefore the function isn't differentiable at $(0,0)$
When to use a hyphen to coin a new word and when to omit a hyphen? Someone has asked for answer to these topics. However, I still want someone to provide me with simple and universal answers. I recently read a sentence from the English-speaking person. It is in an article. It is, ' The available options for mitigating pad cratering are not universally effective (see link below), but they may be helpful on a case by case basis until more crack resistant laminates are developed and become available.  '  If I am to write this, I will use 'a case-by-case basis until more crack-resistant laminates'   Can anyone tell me when to use a hyphen to coin a new word? Let $K$ be a field and $f(x)\in K[x]$. Prove that $K[x]/(f(x))$ is a field if and only if $f(x)$ is irreducible in $K[x]$. Let $K$ be a field and $f(x)\in K[x]$. Prove that $K[x]/(f(x))$ is a field if and only if $f(x)$ is irreducible in $K[x]$.   How to prove? I really have no idea... Thank you a lot. How do I Uninstall Windows 10 or Windows 10 preview? I joined in the  and I don't like it. However, this question may also be relevant to those who install the full version of Windows 10.  How can I downgrade back to Windows 8/7 from Windows 10 or Windows 10 preview? Custom question lists: finding questions you can answer Prepositions used with "inquire" Taking the second derivative of a parametric curve I’ve run into a limitation in the cURL bindings for PHP. It appears there is no easy way to send the same multiple values for the same key for postfields. Most of the workarounds I have come across for this have involved creating the URL encoded post fields by hand tag=foo&amp;tag=bar&amp;tag=baz) instead of using the associative array version of CURLOPT_POSTFIELDS.  It seems like a pretty common thing to need to support so I feel like I must have missed something. Is this really the only way to handle multiple values for the same key?  While this workaround might be considered workable (if not really annoying), my main problem is that I need to be able to do multiple values for the same key and also support file upload. As far as I can tell, file upload more or less requires to use the associate arravy version of CURLOPT_POSTFIELDS. So I feel like I am stuck.  I have  about this problem in more detail on the cURL PHP mailing list in the hopes that someone there has some ideas about this.  Suggestions or hints on where I can look for more information on this are greatly appreciated! Home screen icon for terminal program? I'm using terminal-ide  which is basically a terminal for android with all sorts of utilities. One is able to create bash scripts for example. Is there a way to create an icon on the home screen that will launch one of these scripts? How to split by commas that are not within parentheses? Should I write "X and I", "X and me", "I and X", or "me and X" in a conjoined object? When two accounts are merged, all the posts/votes are moved over to the "master" account, and the other account ("slave"?) is deleted. The merge is propagated network-wide, so a similar thing happens on each site.  But, this does not happen to the chat user. The "slave" chat user is kept, with a 404 parent profile link. I feel that the chat user should be merged as well, re attributing all the chat messages.  Thoughts? If $ a,b,c\in \left(0,\frac{\pi}{2}\right)\;,$ Then prove that $\frac{\sin (a+b+c)}{\sin a+\sin b+\sin c}<1$ If $\displaystyle a,b,c\in \left(0,\frac{\pi}{2}\right)\;,$ Then prove that $\displaystyle \frac{\sin (a+b+c)}{\sin a+\sin b+\sin c}&lt;1$   $\bf{My\; Try::}$ Using $$\sin(a+\underbrace{b+c}) = \sin a\cdot \cos (b+c)+\cos a\cdot \sin (b+c)$$  $$ = \sin a\cdot (\cos b\cos c-\sin b\sin c)+\cos a(\sin b\cos c+\cos b\sin c)$$  $$ = \sin a\cos b\cos c-\sin a\sin b\sin c+\cos a \sin b\cos c+\cos a\cos b\sin c$$  Now how can i solve it after  that , Help required, Thanks Use of "save" or "save for" in sentences How do lexical closures work? While I was investigating a problem I had with lexical closures in Javascript code, I came along this problem in Python:  flist = []  for i in xrange(3):     def func(x): return x * i     flist.append(func)  for f in flist:     print f(2)   Note that this example mindfully avoids lambda. It prints "4 4 4", which is surprising. I'd expect "0 2 4".   This equivalent Perl code does it right:  my @flist = ();  foreach my $i (0 .. 2) {     push(@flist, sub {$i * $_[0]}); }  foreach my $f (@flist) {     print $f-&gt;(2), "\n"; }   "0 2 4" is printed.  Can you please explain the difference ?    Update:   The problem is not with i being global. This displays the same behavior:  flist = []  def outer():     for i in xrange(3):         def inner(x): return x * i         flist.append(inner)  outer() #~ print i   # commented because it causes an error  for f in flist:     print f(2)   As the commented line shows, i is unknown at that point. Still, it prints "4 4 4". Can any one tell me the advantage of synchronized method over synchronized block with an example? Is cross-posting a question on multiple Stack Exchange sites permitted if the question is on-topic for each site? It is possible to migrate a question from one Stack Exchange site to another by closing, but if I have a question that I think is on-topic for multiple Stack Exchange sites, is it OK to post it on both (multipost)?   For example, I have a question that's earned me the tumbleweed badge on SO and I'm not sure what the best thing to do with it is. It's about a web server so it might be answerable on Server Fault but it's really more of a programming thing, hence the posting the question on Stack Overflow.    Is there any way to make the question visible on multiple sites (crosspost) and then accept the answer wherever it came from? Help with proof of $|x+y| \geq |x|-|y|$ WINDOWS 10 - Error Windows Update 10 = Error in phase " SAFE_OS " with the operator error with " BOOT " For days I tried several ways to install Windows 10 . With Windows Update I have always an error and with the media creation tool microsoft I have this message: " The installation failed in SAFE_OS phase with an error when BOOT operator " with this code : 0xC1900101 - 0x20017 Having tried everything I do not know what to do ... Can you help me please ? Thank you  PS: I migrated my OS with the software "Paragon migrate OS to SSD" from my hard disk to my SSD. Maybe I have this errors because I did this.... Is it possible to recover an email that has been clicked "delete forever" in the trash folder? Ackermann function - how to calculate the number of times it calls itself Purely for my own amusement I've been playing around with the .  The Ackermann function is a non primitive recursive function defined on non-negative integers by:  $A(m,n) = n+1$, if $m=0$  $A(m,n) = A(m-1,1)$, if $m&gt;0$ and $n=0$  $A(m,n) = A(m-1,A(m,n-1))$, if $m&gt;0$ and $n&gt;0$  I'm not interested in values of the function itself, rather I'm interested in the number of recursions the function takes to arrive at a value. For example:  $A(1,2) = A(0,A(1,1)) = A(0,A(0,A(1,0))) = A(0,A(0,A(0,1))) = A(0,A(0,2)) = A(0,3) = 4$  That's 6 steps, so if we define $RA(m,n)$ as the number of recursions to calculate $A(m,n)$, then $RA(1,2) = 6$.  I made a short Excel macro to calculate $RA(m,n)$ for any given $m$ and $n$. The problem I have is that $RA(m,n)$ grows even faster than $A(m,n)$ and I quickly run out of stackspace. The table below shows $RA(m,n)$ for very small values of $m$ and $n$. The numbers shown for $RA(3,10)$ and $RA(4,1)$ are the number of steps before crashing!    A formula for $RA(1,n)$ is trivial, and it is relatively easy to see that $RA(2,n) = 5(n+1) + 4n(n+1)/2$.  So my questions: can anyone see an explicit formula for $RA(3,n)$? More generally, does a formula exist for $RA(m,n)$, in the same way that $A(m,n)$ can be expressed using Knuth arrows?   For the low values shown of $m$ and $n$, $RA(m,n)$ looks about order of magnitude larger than $A(m,n)$. Can we prove that $RA(m,n) &gt; A(m,n)$ for all $m,n$. Intuitively, it should be.  For bonus points, can $RA(m,n)$ be expressed in terms of $A(m,n)$?