I've been trying to solve a bunch of these limits by turning to polar coordinates, though I could only deal with the ones that have x and y approaching zero. In this particular case nothing points to this limit being independent of r, and trigonometric functions can't be reduced(easily). I would really appreciate it if you could break down the steps right after applying equivalence and turning to polar coordinates.
121302
$$\lim_{x\to2\  \\ y\to2}\frac{x^{6}+ \tan (x^{2}-y^{2}) - y^{6}}{\sin(x^{6}-y^{6}) - x^{5}y +xy^{5} + \arctan(x^{2}y -xy^{2})}$$ I used a fact that $$\tan \alpha \sim \alpha \\ \arctan \alpha \sim \alpha \\  \sin\alpha \sim \alpha$$ Since now we have $$\lim_{x\to2\ \\ y\to2}\frac{x^{6}+ x^{2}-y^{2} - y^{6}}{x^{6}-y^{6} - x^{5}y +xy^{5} + x^{2}y -xy^{2}}$$ Then $$\lim_{x\to2\ y\to2}\frac{x^{6}+ x^{2}-y^{2} - y^{6}}{x^{6}-y^{6} - x^{5}y +xy^{5} + x^{2}y -xy^{2}}=\\ \\  =\lim_{x\to2\ \\ y\to2}\frac{(x+ y)(x-y)(x^{2}-xy +y^{2})(x^{2}+xy +y^{2})+(x+y)(x-y)}   {(x+ y)(x-y)(x^{2}-xy +y^{2})(x^{2}+xy +y^{2})-xy(x^{2}-y^{2})(x^{2}+y^{2}) - xy(x-y)}=\\=\lim_{x\to2\ \\y\to2}\frac{(x+ y)(x^{2}-xy +y^{2})(x^{2}+xy +y^{2})+(x+y)}   {(x+ y)(x^{2}-xy +y^{2})(x^{2}+xy +y^{2})-xy(x+y)(x^{2}+y^{2}) - xy} $$  And what to do next what multipliers to group? Help please.
30560
I am using extensively trigonometric functions when an angle is given in degrees. Some of these functions like sine or cosine have rational values, for example, the well   known example is that $\cos(\theta) =0.6 $ and $\sin(\theta) =0.8 $.       However besides the case of multiplicity of $90^\circ$ it seems there are no rational numbers $\theta$  with simultaneously rational  values of  sine and cosine.      Is it possible somehow to prove that for rational values of an angle given in degrees there are  no values simultaneously rational of sine and cosine functions, beside obvious case of multiplicities of $90^\circ$?
121743
It's a quite simple question. But I couldn't see it... How to prove that $$\lim_{(x,y) \to (0,0)}\frac{xy^2}{x^2 - y^2}$$ doesn't exist? It's sufficient to show that for different paths through the origin this limit has diferent values (or doesn't exist in some of them). But, what paths to choose?!
91116
In the discussion of linear fractional equations in Birkhoff and Rota's Ordinary Differential Equations, the authors assert that if we convert a DE of the form $y' = F\left(\frac{y}{x}\right)$ to polar coordinates, then we have \begin{align} \frac{1}{r}\frac{dr}{d\theta} = \cot \psi, \end{align} where $\psi = \gamma - \theta$, with $\gamma$ being the tangent direction and $\theta$ the radial direction.  I'm afraid this has me entirely buffaloed -- why on earth is this true?  I have neither an analytic nor geometric intuition as to how this could possibly be.  I'm sure there's some elementaryish fact about $\frac{dr}{d\theta}$ which makes the answer obvious, but I have no clue what said fact is.  I do note that  \begin{align} \frac{dy}{d\theta} &amp;= r\cos\theta + \frac{dr}{d\theta}\sin\theta\Rightarrow\\ \frac{dr}{d\theta} &amp; = \frac{dy}{d\theta}\csc\theta - r\cot\theta\Rightarrow\\ \frac{1}{r}\frac{dr}{d\theta} &amp; = \frac{1}{y}\frac{dy}{d\theta} - \cot\theta,  \end{align} but this is as close as I can come to getting a cotangent anywhere near the expression (and, of course, $\theta \neq \psi$ in general).  What gives?
28228
First I let $x=r\cos \theta, y = r\sin \theta$ and so limit  $$\lim_{r\to 0} r^2\sin2\theta \log(r)$$  Now, in region $0&lt;x&lt;1$, $\log(x) &lt; 1/x$  $$|r^2\sin 2\theta \log (r) - 0| &lt; |r\sin 2\theta| \le |r| &lt; \delta &lt; \epsilon$$  So limit exist if $\delta &lt; \epsilon$ and limit is 0.    Other way, I used L hospital, I don't know if we can apply, but I wrote $r^2 \log r$ as $\log(r) / (r^{-2})$ which again gave 0.
51211
I'm having a difficult time trying to find two different paths that give me different limits  for the following:  $$\lim_{(x,y) \rightarrow (0,0)}  \frac{x-y}{x^2 + y^2}$$
121314
I started resolving this limit:  $$\lim_{x \to 0} \dfrac{\tan 3x}{\tan 5x}$$  so:  $$\begin{align}\lim_{x \to 0} \dfrac{\tan 3x}{ \tan 5x} &amp; = \lim_{x \to 0} \dfrac{\frac{\sin 3x}{\cos 3x }}{\frac{\sin 5x}{\cos 5x}} \\ \\ &amp; = \lim_{x \to 0} \dfrac{ \sin (3x) \cdot \cos (5x)}{ \sin(5x) \cdot \cos (3x)}\\ \\ &amp; = \quad?\end{align}$$  Could someone help me please
18661
How do we get limit such as $0\le\theta\le\pi$, $0\le\phi\le2\pi$ in spherical coordinate system where $$x=r \sin\theta\cos\phi, y=r \sin\theta\sin\phi, z=r \cos\theta$$  Why is the $\theta$-limit $[0,\pi]$ and not $2 \pi$?  I think that the angle of the foot of perpendicular from the point taken in spherical coordinate can lie within $[0,2 \pi]$
117251
I came across this problem when I was at a book store inside of a book made to prepare Berkeley graduates to pass a mandatory exam. I wanted to buy the book, but, alas, I didn't have the money (forty bucks is a lot of money when you don't have a job). So I took my phone out and started taking as many pictures as I could. Unfortunately, I didn't take any pictures of the solutions!  Enough backstory. Time for math. The problem is as follows:  Prove that $\cos^p( \theta) \le \cos(p \theta)$ if $0\le\theta\le\frac\pi2$ and $0\le p\le 1$.  I tried using the series expansion for cosine, but that seemed to be a dead end. Then I tried using Euler's theorem, but I got stuck. Then I got distracted and started to think about other cosine identities. For example, $$\cos\left(\frac\theta2\right)=\pm\sqrt{\frac{1+\cos(\theta)}{2}}.$$ Then what is $\cos(\frac\theta3)$ equal to? I tried to figure it out then I realized that I needed to find the root of a cubic polynomial. Then I realized that I'm better at staying focused than finding the roots of cubic polynomials.  Anyways, a proof would be nice here. I really appreciate any hints or answers. I apologize for my digressions!
2868
In class, we were simply given that this limit is undefined since along the paths $y=\pm x$, the function is undefined.  Am I right to think that this should be the case for any function, where the denominator is $x^2-y^2$, regardless of what the numerator is?  Just wanted to see if this is a quick way to identify limits of this form.  Thanks for the discussion and help!
73606
$\int_{-2}^0\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \sqrt{x^2+y^2} dydx$  y describes the circle of radius 2, and is positive and negative, so doesnt limit any quadrants in our integral, however x can only be negative, which means the left half of our circle is what we are interested in.  This means, we need the integral:  $\int_{\pi/2}^{3\pi/2}\int_{0}^{2} rdrd\theta$  Giving us a value of $2\pi$.  Is this a correct assessment? I wasnt sure if making  $0$ and $2$ the limits of $r$ was entirely correct, since it describes the entire circle of radius $2$, and not just one half. Then again, a circle of radius $2$ does have an area of $4\pi$ which makes me feel I am along the right lines.   Any clarification would be fantastic. Thank you!
98501
What's the value of this limit? I keep getting half, but the answer's 1/3. I think there might be a problem with writing ${x\cot x \over x^2 }$  as $ {\cos x \over x^2}$. Is there ? I get (1-cosx)/x^2 on using this substitution, and if I use L'hôpital's Rule on this, I get the answer as sinx/2x = 1/2.  Edit: I can only use results like $\sin x=x$ and L'Hospital's rule. No series expansion.  Edit 2: Here's my attempt:  1/x^2 - cosx/(x*sinx) = 1/x^2 - (cosx/x^2)(x/sinx) = 1/x^2 - (cosx/x^2)(1)  {Since sinx/x=1} = (1-cosx)/x^2 Being a 0/0 case, L'hôpital's Rule: sinx/2x = (1/2)(sinx/x) = 1/2, since sinx/x=1
112241
So I found this on the Wolfram website today:     So I was wondering about how one might be able to (if possible) solve those four problems by hand. Here are the problems, $\LaTeX$ed:    $ \lim_{x \to +\infty} \dfrac {\sqrt{x^3-x^2+3x}}{\sqrt{x^3}-\sqrt{x^2}+\sqrt{3x}} $ $ \displaystyle\sum_{k=1}^{\infty} \dfrac {(-1)^{k+1} k^2}{k^3+1} $  $ \dfrac {\mathrm{d}}{\mathrm{d}u} \left[ \dfrac {u^{n+1}}{(n+1)^2} \cdot \left[ (n+1) \ln u - 1 \right] \right] $  $ \displaystyle\int_0^{2\pi}\displaystyle\int_0^{\frac{\pi}{4}}\displaystyle\int_0^4 \left( \rho \cos \phi \right) \rho^2 \sin \phi \, \mathrm{d}\rho \mathrm{d}\phi \mathrm{d}\theta$     Ideas   The degree of the numerator is $\frac{3}{2}$. The degree of the denominator is $\frac{3}{2}$. The expression is of an indeterminate form, namely $\frac{\infty}{\infty}$, so we use l'Hoptial's Rule.  No ideas, really.  Derivatives are always pretty easy, although this one is a bit bashy. Basically bash Product and Chain Rules, etc.  Integration by Parts bash? The $\mathrm{d}\theta$ part is very trivial. I mean for the $\rho$ and $\phi$.
66671
So I got this assignment. And I was wondering how is it possible to get a limit from a constantly changing formula.  $$\lim_{x\to+\infty}\frac{\sin(x)}{x}$$ Can I only look in the domain $]0,2\pi[$?
54519
I am not being able to understand the graphical method of solving this, any simple explanation will be appreciated.  A non-graphical calculation will be very helpful too.  Thank you so much in advance!
44763
I have been struggling to solve this limit.  What is the limit as $x$ approaches $45^0$ of $$\frac{\sqrt{2}\cos x-1}{\cot x-1}?$$   I know how to use L'Hospital's rule to calculate this limit and got the answer as $0.5$. But, how do I calculate the limit by manipulating the function? Please provide only some hints to proceed.
59001
How to show if the limit $$ \lim_{(x,y) \to (0,0)} \frac{x^3}{x^2 + y^2}$$ exists? I suspect that there is, as I can't find any path that would show that limit doesn't exist, and WolframAlpha also suggests that the limit is (0,0).  In general, can you recommend any tips how to learn to approach similar limit problems (fractions and polynomials like this)? edit That is, excluding the polar coordinate conversion method?