Recent content by laura1231

$$\displaystyle\lim_{(x,y)\rightarrow(x_0,0)}(x^2+y^2)\arctan\dfrac{1}{|xy|}=$$ $$=\displaystyle\lim_{r\rightarrow0}[(x_0+r\cos\phi)^2+(r\sin\phi)^2]\arctan\dfrac{1}{|(x_0+r\cos\phi)r\sin\phi|}=$$...

Can I extend the function $f(x,y)=(x^2+y^2)\arctan\dfrac{1}{|xy|}$ to a continuous function? If I consider the restriction of $f$ along the line $x=k$ i find $\lim_{(x,y)\rightarrow(k,0)}(x^2+y^2)\arctan\dfrac{1}{|xy|}=k^2\dfrac{\pi}{2}$ how can i prove that?

We can also solve it in this way: $X=\cos x$ $Y=\sin x$ the equation $(2-\sqrt{2})(1+\cos x)+\tan x=0$ becomes: $(2-\sqrt{2})(1+X)+\dfrac{Y}{X}=0$. From $\cos^2 x+\sin^2 x=1$ we have $X^2+Y^2=1$, hence: $$ \left\{\begin{array}[l] ((2-\sqrt{2})(1+X)+\dfrac{Y}{X}=0\\X^2+Y^2=1\end{array}\right...

Thanks! There is another solution. When you use double-angle identities for cosine and tangent you have $x\neq \pi+2k\pi$, but this is also a solution of the equation.

Hi, I've tried to solve this equation: $(2-\sqrt{2})(1+\cos x)+\tan x=0$ and I've tried everything but nothing works...Does anybody have an idea?

I try:

$\displaystyle\dfrac{2x-3}{x^2-x}dx=\dfrac{dy}{y}\Rightarrow\ \int\dfrac{2x-3}{x^2-x}dx=\int\dfrac{dy}{y}$ $y=\dfrac{kx^3}{x-1}$

Hi, in a book I have found this problem: "Let be $f,g:\mathbb{R}\rightarrow\mathbb{R}$ two derivable functions such that $f(0)=g(0)$ and $f(6)=g(6)$. Which of the following statements is necessarily true?: a) $\exists\ c\in]0;6[ : f'(c)=g'(c)$; b) $\exists\ c_1,c_2\in]0;6[ : f'(c_1)=g'(c_2)$. "...

Thanks $\chi\ \sigma$, in effect this equation is very hard to solve... I think that is impossible to find an analytic solution.. The numerical solution seems to be the only way at moment...

yes $\chi\sigma$, it seems simple...apparently

thanks for your answer, this integral is a consequence of an attempt to solve this differential equation $y''+y-\dfrac{y}{1+x^3}=0$...

Hi, I tried to solve this integral $$\int\sqrt{1-\frac{1}{x^3}}dx$$ but i can't solve it... can someone help me?

Re: triple integral Thanks $\chi\sigma$ but the problem is in $\int z^4 e^{z^4}dz$...

Hi! I have some problems with the integral $$\iiint_T\sqrt{x^2+y^2}z^4e^{z^4}dx\ dy\ dz$$ where $$T=\{(x,y,z)\in\mathbb{R}^3:\sqrt{x^2+y^2}\leq z\leq 1\}$$ I have tried to change it to spherical and cylindrical coordinates but... nothing Can someone help me? Thanks