Recent content by Francolino

Can anybody help me out with this proof?

Yes, so in my case $ (a,b) = (0,0) $ then, I have to show that: $$ \sqrt{x^2+y^2} =: \left \| (x,y) \right \| < \delta \Rightarrow \left |\frac {e^{x+y^2}-1-\sin \left ( x + \frac{y^2}{2} \right )}{x^2+y^2} - \frac{1}{2} \right | < \varepsilon $$ And that was exactly what I was trying to show.

You're right! I was reading other limit and I confused the result (the other limit was zero). I've changed the proof: $$ \begin{align*} \left |\frac {e^{x+y^2}-1-\sin \left ( x + \frac{y^2}{2} \right )}{x^2+y^2} - \frac{1}{2} \right | &\leq \frac {\left |e^{x+y^2}-1-\sin \left ( x +...

What I've tried so far: $$ \begin{align*} \left |\frac {e^{x+y^2}-1-\sin \left ( x + \frac{y^2}{2} \right )}{x^2+y^2} \right | &\leq \frac {\left |e^{x+y^2}-1-\sin \left ( x + \frac{y^2}{2} \right ) \right |}{\left \| (x,y) \right \|^2} \\ &\leq \frac {\left |e^{x+y^2}-1 \right |+ \left...

Well, I tried to do something similar to what I was suggested to do in http://mathhelpboards.com/calculus-10/convergence-sequence-15868.html. So I took polar coordinates: Using that: $$ \left\{\begin{matrix} a_{n} = r_{n}\cos(\theta_{n}) \\ b_{n} = r_{n}\sin(\theta_{n}) \end{matrix}\right. $$...

Well, clearly the rotation won't decide the convergence. Then, if $ r_{n} \geq 1 $ it'll diverge. So $ r_{n} < 1 $ it'll converge (and we don't care about $ \theta_{n} $?). So if $ r_{n} < 1 $ then $ r_{n}^2 < 1 $. Now, squaring and summing both equations (the ones I used to convert into polar...

I came up to this: $$ \left\{\begin{matrix} r_{n+1}\cos(\theta_{n+1}) = r_{n}\cos(\theta_{n} + \pi/3) \\ r_{n+1}\sin(\theta_{n+1}) = r_{n}\sin(\theta_{n} + \pi/3) \end{matrix}\right. $$ (I hope I didn't make any silly mistake).

I must be blind, but I cannot see it. Anyway, the fact that $ \cos(\pi/3) = 1/2 $ and $ \sin(\pi/3) = \sqrt{3}/2 $ seems very curious to me. Is $ \pi/3 $ important in this context?

First of all, thank you for answering! I did what you suggested: $$ \left\{\begin{matrix} \displaystyle r_{n+1}\cos(\theta_{n+1}) = r_{n}\left ( \frac{1}{2}r_{n}\cos(\theta_{n}) - \frac{\sqrt{3}}{2}r_{n}\sin(\theta_{n}) \right )\ \\ \displaystyle r_{n+1}\sin(\theta_{n+1}) = r_{n}\left (...

I literally don't know how to solve this one. I hope you can help me. :)

I apologize myself. I commited a mistake typing my last message. Now, the error is fixed. Forget about $ n $. It's $ x $, now. Sorry.

Thanks again, for answering. :) $$\displaystyle \lim_{x \to \infty} \sin (x\pi\sqrt [3] {x^3+3x^2+4x-5} + 0) = \displaystyle \lim_{x \to \infty} \sin (x\pi\sqrt [3] {x^3+3x^2+4x-5} + [x·(x+1)·\pi - x·(x+1)·\pi])$$ As $x·(x+1)·\pi$ is always a pair number (no matter what intenger $x$ is), so...

Hi to everyone. I'm new on here (in fact, this is my really first message). I need some help with the next limit, I hope you can help me: $$\lim_{x \to \infty} \sin (x\pi\sqrt [3] {x^3+3x^2+4x-5})$$ Thank you so much for your time! :)