If a function $f(x) = x^2 \sin (\frac{1}{x}), x \neq 0$ and $f(0) = 0$, how do I show this function is differentiable on $\mathbb{R}.$  --Edit  I have tried to find the limit when $x \neq 0$ and got this. The definition of the limit of a function is $f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}.$ So I got $f'(a) = \lim_{h \rightarrow0} \frac{a^2 (\sin(\frac{1}{a+h}) - \sin(\frac{1}{a}))}{h} = 0$ at $x = a,$ because as $h \rightarrow0$, $\sin(\frac{1}{a+h}) = \sin(\frac{1}{a}).$ Am I doing it right??
Let $f(x) = x^2 \sin{\frac{1}{x}}$ for $x\neq 0$ and $f(0) =0$. (a) Use the basic properties of the derivative, and the Chain Rule to show that $f$ is differentiable at each $a\neq 0$ and calculate $f'(a)$. You may use without proof that $\sin$ is differentiable and that $\sin' =\cos$.  Not even sure what this is asking.  (b) Show that $f$ is differentiable at $0$ and that $f'(0) =0$.  $\frac {f(x)-f(0)}{x-0} \to \lim_{x \to 0} x \sin(1/x)$. $x \sin(1/x) \leq |x|$ and $\lim_{x \to 0} |x|=0$. Thus $f(x)$ is differentiable at $0$; moreover $f^{'}(0)=0$.  (c) Show that $f'$ is not continuous at $0$.  $f{'}(x)=x^{2} \cos(1/x) (-x^{-2}) + 2x \sin (1/x)$. In pieces: $\lim_{x \to 0} \cos (1/x)$. $f^{'}(0-)$ nor $f{'}(0+)$ exists as $x \to 0$ $f^{'}(x)$ oscillates infinity between $-1$ and $1$ with ever increase frequency as $x \rightarrow 0$ for any $p&gt;0$ $[-p,0]$, $[-p,p]$ or $[0,p]$ $f$ is not continuous. Question: How to show more rigorously?
Why is my Mac constantly trying to connect to another server? After logging on, I see this dialog box pop up. Whenever I dismiss it, it reappears 30 seconds later. (Stealing focus at the same time! Grrrr!)    "Slate" is my old iMac, which is sitting on the desk next to me, turned off. How can I stop my new Mac from constantly trying to connect to my old one? Both are running 10.7.1. If $\sum{a_n}$ converges does that imply that $\sum{\frac{a_n}{n}}$ converges? I know if $\sum{a_n}$ converges absolutely then $\sum{\frac{a_n}{n}}$ converges since $0\le \frac{|a_n|}{n} \le |a_n| $ for all $n$ so it converges absolutely by the basic comparison test and therefore converges. However, I cannot prove the convergence of $\sum \frac{a_n}{n}$ if $\sum{a_n}$ converges but not absolutely even though I suspect it to be true. Can you give me a proof or a counterexample for this? Can you nest html forms? Using wp_query is it possible to orderby taxonomy? My question is simple, I'm using WP_Query to retrieve some custom type posts filtering by a taxonomy using tax_query.  Now my problem is I would like to orderby the taxonomy, but from documentation and searching on the web I can't find a solution.  The orderby in WP_Query lets you order by a bunch of fields even custom meta fields but it doesn't seem to support taxonomy.  Any pointers in the right direction?  Thank you all. Prove $GL_n(\Bbb{R})/SL_n(\Bbb{R}) \cong \Bbb{R}^\times$ $$GL_n(\Bbb{R})/SL_n(\Bbb{R}) \cong \Bbb{R}^\times$$  This is trivial to prove with the first isomorphism theorem - by using the determinant as a endomorphism, then as $SL_n(\Bbb{R})$ is $1$ under the determinant, it is the kernel, and by FIT, the above is proved.  I was wondering how to prove this without the isomorphism theorem ? How to vertically center content with variable height within a div? Which one is correct?     There is an apple and an orange..   or     There are an apple and an orange?" Finding the variable name passed to a function pure code formatting, to enhance a question, is impossible (without 2k rep) Many To many for Bulk records I have created two custom objects (A and B)and a junction object(AB). i.e. Both A and B have Many to Many relationship. Now I need to insert 5000 records. How can I automatically populate the relation in Junction object. DO I have to write any trigger or Do I have to do Manually mapping the records? Proving $\arg(zw)=\arg(z)+\arg(w)$ Background Nowadays, I see &quot;them&quot; used to mean &quot;those&quot; a lot. I don't know if it was as common in the past. For example, take &quot;one of &quot;. On researching about it, I found some people say it comes from a dialect of British English. Some others say it is a &quot;non-standard&quot; usage. I see this usage in Canadian English also, and it seems some people use it in a sarcastic way. Moreover, I have seen a song titled &quot;one of them days&quot;. I also read, in the book called &quot;A Broken Promise&quot;, &quot;&quot; Finally, Wikipedia says that it is a usage in  (a common name for the Southern Midland dialect of American English):  Pronouns and demonstratives &quot;Them&quot; is sometimes used in place of &quot;those&quot; as a demonstrative in both nominative and oblique constructions. Examples are &quot;Them are the pants I want&quot; and &quot;Give me some of them crackers.&quot;    Question(s): What would you say about the usage of this word? Is it correct? Could we use it in daily speech? Could this usage go beyond a specific dialect and be used in other dialects, regions, etc.? Does it really originate from Appalachian English? Why did this usage become popular among other English speakers? Note: I have already seen a similar question:  However, it only says, &quot;ungrammatical,&quot; there. This question is specific to this situation only, and there is more to it. Chen weibull distribution Background I updated my TeX Live installation on my Mac a while ago and got the error LaTeX3 Error: Mismatched LaTeX support files detected. I found a solution at .  As far as I understand it (which isn't very far), there are system formats and local formats, and TeX Live is updating the wrong one.  The fix is to update the formats. Using fmtutil-user --all worked for me.  It spewed out a load of stuff, then output fmtutil [INFO]: log file copied to: /Users/&lt;user&gt;/Library/texlive/2020/texmf-var/web2c/luatex/dvilualatex.log fmtutil [INFO]: /Users/&lt;user&gt;/Library/texlive/2020/texmf-var/web2c/luatex/dvilualatex.fmt installed.  I don't know what this tells me, but my TeX Live preferences has TeX programs set at /Library/TeX/texbin, which seems to be a link to /usr/local/texlive/2020/bin. Problem Every time I update TeX Live, I have to apply the solution again. Question Is there a way to make TeX Live use the correct formats permanently so that I don't have to keep running fmtutil-user --all after every TeX Live update?  Extra information  to  check if you have some local formats around which hide the system formats. (with eg kpsewhich --all --engine=pdftex pdflatex.fmt)  When I did that, it output /Users/&lt;user&gt;/Library/texlive/2020/texmf-var/web2c/pdftex/pdflatex.fmt /usr/local/texlive/2020/texmf-var/web2c/pdftex/pdflatex.fmt  but I don't know what that's telling me. From Berkeley Problems in Mathematics, Spring 1999, .  Let $f(x)\in \mathbb{Q}[x]$ be an irreducible polynomial of degree $n\ge 3$. Let $L$ be the splitting field of $f$, and let $\alpha\in L$ be a zero of $f$. Give that $[L:\mathbb{Q}]=n!$, prove that $\mathbb{Q}(\alpha^{4})=\mathbb{Q}(\alpha)$.   (BELOW IS NONSENSE, SHOULD BE IGNORED)  Following Gerry Myerson's advice, since $L$ is the splitting field of $f$, $L$ must be Galois over $\mathbb{Q}$. Thus the Galois group for $[L:\mathbb{Q}]=S_{n}$. Gerry now asserts that either $\mathbb{Q}(\alpha^{4})$ is of degree 2 over rationals or it must be $\mathbb{Q}$. But assuming this one can rule out the case $\mathbb{Q}$, since this would imply $\alpha$ is the root of a 4th or 2nd degree polynomial over $\mathbb{Q}$. But we know $n\ge 3$. So $\alpha$ must be the root of a 4th degree polynomial. We know $S_{n}$ has normal subgroups $A_{n}$ when $n\not=4$ and $V,A_{4}$ when $n=4$. I do not know how to proceed any further.   Now assuming $\mathbb{Q}(\alpha^{4})$ is a degree 2 abelian extension over $\mathbb{Q}$. We should have a chain of normal extensions $\mathbb{Q}(\alpha)\supset \mathbb{Q}(\alpha^{2})\supset \mathbb{Q}$. This would imply $S_{n}$ has a normal subgroup which has a normal subgroup. But we know $A_{n}$ for $n\ge 5$ are simple. So the only possibility is $\mathbb{Q}(\alpha)$ has a Galois group isomorphic to $V$. In this case $n=4$ as well. I can only proceed to here. Lower case "k" in Cocoa I know this is a common convention, but what does the "k" in variable names signify? (i.e. kMaxImageViewSize)  I looked in the Apple documentation on Variable names and found no mention of it.  Thanks for answering Is RSA key size the size of private key exponent? I have implemented a key pair generation scheme for RSA algorithm. I have taken the length of private key exponent as RSA key size, but then I've got to know that RSA key size is the size of the modulus $n$. Is it wrong to take RSA key size as the size of the private key exponent?  To generate a private key of 1024 bits, I increase the size of private key exponent regardless the modulus $n$. How to tell if a  tag failed to load I'm dynamically adding &lt;script&gt; tags to a page's &lt;head&gt;, and I'd like to be able to tell whether the loading failed in some way -- a 404, a script error in the loaded script, whatever.  In Firefox, this works:  var script_tag = document.createElement('script'); script_tag.setAttribute('type', 'text/javascript'); script_tag.setAttribute('src', 'http://fail.org/nonexistant.js'); script_tag.onerror = function() { alert("Loading failed!"); } document.getElementsByTagName('head')[0].appendChild(script_tag);   However, this doesn't work in IE or Safari.  Does anyone know of a way to make this work in browsers other than Firefox?  (I don't think a solution that requires placing special code within the .js files is a good one. It's inelegant and inflexible.) How do I split a string, breaking at a particular character? I have this string  'john smith~123 Street~Apt 4~New York~NY~12345'   Using JavaScript, what is the fastest way to parse this into  var name = "john smith"; var street= "123 Street"; //etc... Using Integration By Parts results in 0 = 1 I've run into a strange situation while trying to apply Integration By Parts, and I can't seem to come up with an explanation.  I start with the following equation:  $$\int \frac{1}{f} \frac{df}{dx} dx$$  I let:  $$u = \frac{1}{f} \text{ and } dv = \frac{df}{dx} dx$$  Then I find:  $$du = -\frac{1}{f^2} \frac{df}{dx} dx \text{ and } v = f$$  I can then substitute into the usual IBP formula:  $$\int udv = uv - \int v du$$  $$\int \frac{1}{f} \frac{df}{dx} dx = \frac{1}{f} f - \int f \left(-\frac{1}{f^2} \frac{df}{dx}\right) dx$$  $$\int \frac{1}{f} \frac{df}{dx} dx = 1 + \int \frac{1}{f} \frac{df}{dx} dx$$  Then subtracting the integral from both sides, I've now shown that:  $$0 = 1$$  Obviously there must be a problem in my derivation here...  What wrong assumption have I made, or what error have I made?  I'm baffled.