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Since $\sin x$ is a continuous function at $(0,0)$ it suffices to check if the limit $\lim_{(x,y)\to (0,0)} \frac{x^2}{x+y}$ is finite.

I seem to be missing the idea in order to show that the limit $\lim_{(x,y)\to (0,0)} \frac{x^2}{x+y}=0$. I tried converting to polar form I get $\lim_{r\to 0} \frac{r \cos^2 \theta}{\cos \theta +\sin \theta}$. It seems that this is not conclusive. Any help?

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  • $\begingroup$ Are you sure that $lim_{(x,y) \to (0,0)} \frac{x^2}{x+y}$ exists? Because when you approach origin from $t \to 0$ with $x=t$ and $y = -t$ it becomes undefined. $\endgroup$ – crbah Feb 26 '16 at 9:16
  • $\begingroup$ For $y(x)=-x+x^2/l$, $\exists \lim_{x \to 0}y(x)=0, \exists \lim_{x \to 0}{x^2 \over x+y(x)}=l$. $\endgroup$ – Abstraction Feb 26 '16 at 9:22
  • $\begingroup$ @cbahadir I added the extra condition in the domain. $\endgroup$ – Miz Feb 26 '16 at 9:22
  • $\begingroup$ @Miz You don't need that extra condition: it must be that way otherwise you don't have a valid mathematical expression. $\endgroup$ – DonAntonio Feb 26 '16 at 9:23
  • $\begingroup$ @Joanpemo $\lim_{h \to 0} \frac{h}{0}$ is a valid mathematical expression. $\endgroup$ – crbah Feb 26 '16 at 9:27
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We can see there's a problem with your method if

$\;\cos\theta+\sin\theta=0\iff\tan\theta=-1\iff \theta=-\frac\pi4\;$ , in the trigonometric circle.

We can try for example:

$$\begin{align*}&y=x:\implies \frac{x^2}{x+y}=\frac{x^2}{2x}=\frac x2\xrightarrow[(x,y)\to(0,0)]{}0\\{}\\ &y=x^2-x:\implies\frac{x^2}{x+y}=\frac{x^2}{x+x^2-x}=\frac {x^2}{x^2}=1\xrightarrow[(x,y)\to(0,0)]{}1\end{align*}$$

and thus the limit doesn't exist.

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In fact, along the curve $y=x^3-x,$ $\sin(x^2/(y+x)) = \sin(1/x)$ takes on every value in $[-1,1]$ infinitely many times in any deleted neighborhood of $(0,0).$

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For $k\ne 0$ and $0\ne x\ne k,$ let $y_{(k,x)}=-x+x^2/k$. Then $y_{(k,x)}\ne 0$ and $x^2/(x+y_{(k,x)})=k.$ Obviously $\lim_{x\to 0}\sin (\;x^2/(x+y_{(k,x)})\;)=\sin k$.

Also we have $y_{(k,x)}\to 0$ as $x\to 0.$

For example with $k=\pi$ we have $\sin (\;x^2/(x+y_{(\pi,x))}\;)\to 0$ as $x\to 0.$ With $k=\pi /2$ we have $\sin (\;x^2/(x+y_{(\pi /2,x)}\;)\to 1$ as $x\to 0.$

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