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I have tried to use polar coordinates, which gives me:

$$\frac{\sin(r^2)}{1-\cos r}$$

And from here, I couldn't proceed no matter what I tried and then I tried the following identities In a bit of despair: The first is merely the fundamental limit, the second may be derived from it, I guess.

$$\sin x \approx x \quad \quad \quad \quad \quad \quad \quad \quad \cos x \approx 1-\frac{x^2}{2}$$

Which gives me:

$$\frac{r^2}{1-(1-\frac{r^2}{2})}=\frac{r^2}{\frac{r^2}{2}}=\frac{2r^2}{r^2}$$

And then:

$$\lim_{r\to0}\frac{2r^2}{r^2}=2 $$

The same result would also follow without the use of polar coordinates:

$$\frac{x^2+y^2}{\frac{x^2+y^2}{2}}= \frac{2(x^2+y^2)}{(x^2+y^2)}$$

And hence:

$$\lim_{(x,y)\to 0} \frac{2(x^2+y^2)}{(x^2+y^2)}=2$$

The result to both limits was confirmed by Wolfram Alpha and I see no other way to do it. Notice that it's highly "suspicious" that there is a square root inside the $\cos$ and there is a square in our identity for $\cos x$.

$$\Large\text{ Questions:}$$

  1. Is it correct? I feel as if I had done some forbidden move in there.
  2. How can I deduce $\cos x \approx 1-\frac{x^2}{2}$? I know it may follow from the fundamental limit, but I have tried a few things a nothing worked.
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    $\begingroup$ $\cos x=1-x^2/2+O(x^4)$ via the Maclaurin series. if you don't like that, $1-\cos r=2\sin^2(r/2)$. $\endgroup$ – Angina Seng Aug 31 '18 at 16:12
  • $\begingroup$ I think, since the converted fraction into polar coordinates doesn't depend to $\theta$, so the limit doesn't existt. $\endgroup$ – mrs Aug 31 '18 at 16:17
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    $\begingroup$ The square root is not suspicious, it is the square root that computes the modulus of $(x,y)$. The square (of the square root) under the sine is more artificial. $\endgroup$ – Yves Daoust Aug 31 '18 at 16:26
  • $\begingroup$ @BillyRubina Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… $\endgroup$ – user Sep 17 '18 at 20:31
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We don't need polar coordinates indeed it suffices observe that $t=x^2+y^2 \to 0$ and then the limit becomes

$$\lim_{t\to 0}\frac{\sin(t)}{1-\cos \sqrt{t}}=\lim_{t\to 0}\frac{\sin(t)}{t}\frac{t}{1-\cos \sqrt{t}}=2$$

For the standard limit we can also use l'Hopital to obtain

$$\lim_{t\to 0}\frac{1-\cos t}{t^2}=\lim_{t\to 0}\frac{\sin t}{2t}=\frac12$$

or as an alternative

$$\frac{1-\cos t}{t^2}=\frac{1-(1-2\sin^2(t/2))}{t^2}=\frac12\frac{\sin^2(t/2))}{(t/2)^2}\to \frac12$$

Your first way by polar coordinates is not a wrong idea and you can conlude by standard limits as shown here. Note that if we want use series expansion we need to proceed as follow using little-o notation for the remainder (or also big-O notation if you prefer)

$$\frac{\sin(r^2)}{1-\cos r}=\frac{r^2+o(r^2)}{1-\left(1-\frac12r^2+o(r^2)\right)}=\frac{r^2+o(r^2)}{\frac12r^2+o(r^2)}=\frac{1+o(1)}{\frac12+o(1)}\to 2$$

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I would write $$\frac{\sin(r^2)(1+\cos(r))}{(1-\cos(r))(1+\cos(r))}$$

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The bidimensional limit

$$\lim_{(x,y)\to0}f\left(\sqrt{x^2+y^2}\right)$$ can very well be replaced by a unidimensional one

$$\lim_{r\to0}f(r).$$

Specifically,

$$\lim_{r\to0}\frac{\sin r^2}{1-\cos r}=\lim_{r\to0}\frac{\sin r^2}{2\sin^2\dfrac r2}=2\lim_{r\to0}\frac{\sin r^2}{r^2}\left(\frac{\dfrac r2}{\sin\dfrac r2}\right)^2=2.$$

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    $\begingroup$ No Taylor, no L'Hospital, no Wolfram. $\endgroup$ – Yves Daoust Aug 31 '18 at 16:35
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$\dfrac{\sin r^2}{1-\cos r} = \dfrac{\sin r^2}{r^2} \dfrac{r^2}{1-\cos r}$

Using L'Hospital's rule, $\dfrac{r^2}{1-\cos r} \to \dfrac{2r}{\sin r} \to 2$ as $r \to 0$.

So $\dfrac{\sin r^2}{1-\cos r} \to 2$ as $r \to 0$.

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