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

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

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 $1+\rho\cos(\varphi)+(\rho\sin(\varphi))^2$.

• Can you show the work you did in polar coordinates to get that the limit doesn't exist? Jul 11, 2015 at 8:20
• I just added it. :) Jul 11, 2015 at 8:38
• Unfortunately you provide no justification whatsoever for the last $=$ sign in the post, which at present remains a complete mystery.
– Did
Jul 11, 2015 at 8:44
• I already clarified the question. Jul 11, 2015 at 8:49
• ("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, 2015 at 8:50

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 $1/2.$

• Feb 29, 2016 at 7:00
• @KeithMcClary Not sure what your comment means.
– zhw.
Feb 29, 2016 at 16:06