4

$$ \lim_{(x,y)\to(0,0)} \frac {e^{x+y^2}-1-\sin \left ( x + \frac{y^2}{2} \right )}{x^2+y^2} $$

I've a few doubts about this limit. I mean, if I take polar coordinates, I get that the limit doesn't exist. And Wolfram agrees with me. Even though, I've found a solution of this problem that doesn't say the same thing; which I transcribe next:

· The Taylor's second order polynomial of $ e^{x+y^2} $ in $ (0,0) $ is $ 1 + x + \frac{1}{2}x^2 + y^2 $.

· The Taylor's second order polynomial of $ \sin \left ( x + \frac{y^2}{2} \right ) $ in $ (0,0) $ is $ x + \frac{1}{2}y^2 $.

Then:

$$ \lim_{(x,y)\to(0,0)} \frac {e^{x+y^2}-1-\sin \left ( x + \frac{y^2}{2} \right )}{x^2+y^2} = \lim_{(x,y)\to(0,0)} \frac {1 + x + \frac{1}{2}x^2 + y^2-1-(x + \frac{1}{2}y^2)}{x^2+y^2} = \frac {1}{2} $$

Where's the mistake in this?

What I've tried so far:

$$ \lim_{(x,y)\to(0,0)} \frac {e^{x+y^2}-1-\sin \left ( x + \frac{y^2}{2} \right )}{x^2+y^2} $$

Let be $ \rho \geq 0 $ and $ \varphi \in [0,2\pi) $, so:

$$ \left\{\begin{matrix} x = \rho\cos(\varphi)\\ y = \rho\sin(\varphi) \end{matrix}\right. $$

Then:

$$ \begin{align*} \lim_{\rho\to 0} \frac {e^{\rho\cos(\varphi)+(\rho\sin(\varphi))^2}-1-\sin \left ( \rho\cos(\varphi) + \frac{(\rho\sin(\varphi))^2}{2} \right )}{(\rho\cos(\varphi))^2+(\rho\sin(\varphi))^2} &= \lim_{\rho\to 0} \frac {\rho\cos(\varphi)+(\rho\sin(\varphi))^2-\left ( \rho\cos(\varphi) + \frac{(\rho\sin(\varphi))^2}{2} \right )}{(\rho\cos(\varphi))^2+(\rho\sin(\varphi))^2} \\ &= \lim_{\rho\to 0} \frac {\rho\cos(\varphi)+(\rho\sin(\varphi))^2-\left ( \rho\cos(\varphi) + \frac{(\rho\sin(\varphi))^2}{2} \right )}{\rho^2} \\ &= \lim_{\rho\to 0} \frac {(\rho\sin(\varphi))^2- \frac{(\rho\sin(\varphi))^2}{2}}{\rho^2} \\ &= \lim_{\rho\to 0} \frac {1}{2}\sin^2(\varphi) \end{align*} $$

Which limit doesn't exit, because the result depends of $ \varphi $ which varies in $ [0,2\pi) $.

And here you can see that Wolfram agrees.

According to a comment below, I've calculated a third order Taylor's polynomial (respecto to $ \rho$) of $ e^{\rho\cos(\varphi)+(\rho\sin(\varphi))^2} $. I found that any Taylor's polynomial (in $ \rho = 0 $) of order grater than 2 is exactly $ 1+\rho\cos(\varphi)+(\rho\sin(\varphi))^2 $.

francolino
  • 379
  • 2
    Can you show the work you did in polar coordinates to get that the limit doesn't exist? – JimmyK4542 Jul 11 '15 at 08:20
  • I just added it. :) – francolino Jul 11 '15 at 08:38
  • Unfortunately you provide no justification whatsoever for the last $=$ sign in the post, which at present remains a complete mystery. – Did Jul 11 '15 at 08:44
  • I already clarified the question. – francolino Jul 11 '15 at 08:49
  • 1
    ("Already"? No, after my comment--but it is good that you did.) And now the mistake is clear: why stop the expansion of $e^{\rho\cos(\varphi)+(\rho\sin(\varphi))^2}$ at $1+\rho\cos(\varphi)+(\rho\sin(\varphi))^2$? – Did Jul 11 '15 at 08:50
  • Yes, I wrote "already" because I was talking in present, I corrected the post by then. First I corrected it, then I commented that. And I don't understand why should I write a third order Taylor's polynomial. I mean, the greatest terms don't cancel each other. And besides, why Wolfram says that the limit doesn't exist? – francolino Jul 11 '15 at 08:57
  • @francolino, your limit is $1/2$, I think – Michael Galuza Jul 11 '15 at 10:40
  • Thank you for the answer. Where do you think is the mistake taking polar coordinates? – francolino Jul 11 '15 at 10:49
  • The mistake (already pointed out by Did) is that you stopped the expansion of $e^{\rho \cos \varphi + (\rho \sin \varphi)^2}$ at $1+ \rho \cos \varphi + (\rho \sin \varphi)^2$. The next term would have been $\tfrac{1}{2}[\rho \cos \varphi + (\rho \sin \varphi)^2]^2$ which when expanded contains a $\tfrac{1}{2}\rho^2\cos^2\varphi$ term. Add this to the $\tfrac{1}{2}\sin^2\varphi$ term to get $\tfrac{1}{2}$. – JimmyK4542 Jul 11 '15 at 22:26
  • Math question: what @JimmyK4542 said. WA question: dunno, the thing is known to generate countless bugs, to receive a more specific answer you might want to ask the question on a site dedicated to WA. – Did Jul 11 '15 at 23:03

1 Answers1

1

For $(x,y)$ near $(0,0),$ we have

$$e^{x+y^2} = 1 + (x+y^2) +(x+y^2)^2/2 + O((x+y^2)^3), \sin (x+y^2/2) = (x+y^2/2) +O((x+y^2/2)^3).$$

So the numerator of our expression is

$$(y^2/2 + x^2/2) + xy^2 + y^4/2 + O((x+y^2)^3) + O((x+y^2/2)^3)= (y^2/2 + x^2/2) + r(x,y).$$

Dividing by $x^2 + y^2$ we get $1/2 + r(x,y)/(x^2+y^2).$ Now $r(x,y)/(x^2+y^2)\to 0.$ I'll leave this to you for now. Thus the desired limit is $1/2.$

zhw.
  • 105,693