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Limit of $$\lim_{(x,y)\rightarrow(0,0)}\frac{\sin(2x+2y)-2x-2y}{\root\of {{x^2+y^2}}}$$

How can evaluate this limit?
I tried using polar coordinates like $x=r\cos\theta$ and $y=r\sin\theta$. put the thing in the sin doesn't turn into something neat.
Also i tried evaluating along different paths. But I don't know how that helps.

Edit: Also could you prove that the limit you found is the actual limit using the $\epsilon - \delta$ method.

Thanks!

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  • $\begingroup$ Firstly we need to have a rough idea of what it is going on. Note the the numerator is in the form $\sin t - t \sim t^3/6$ with $t=r\cdot 2(\cos \theta+2\sin \theta)$ and the denominator is $r$. $\endgroup$ – user Sep 23 '18 at 6:23
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HINT

We have that

$$\frac{\sin(2x+2y)-2x-2y}{\root\of {{x^2+y^2}}}=\frac{2x+2y}{\root\of {{x^2+y^2}}}\cdot \left(\frac{\sin(2x+2y)}{2x+2y}-1\right)$$

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  • $\begingroup$ +1 Thanks for the answer!, After that do we need to calculate the limit separately for the two terms and then multiply them? But I think the limit for $\frac{2x+2y}{\root\of {x^2+y^2}}$ doesn't exist? $\endgroup$ – SmarthBansal Sep 23 '18 at 6:22
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    $\begingroup$ @SmarthBansal It suffices to show that LHS term is bounded. $\endgroup$ – user Sep 23 '18 at 6:24
  • $\begingroup$ @SmarthBansal for the RHS term just take $2x+2y=t \to 0$ as variable. $\endgroup$ – user Sep 23 '18 at 6:24
  • $\begingroup$ Got ya, nice solution. Last thing, If I wanted to prove that the limit is zero using $\epsilon- \delta$ method how would that go about? $\endgroup$ – SmarthBansal Sep 23 '18 at 6:26
  • $\begingroup$ The same manipulation ca be useful to reduce to prove that $\left(\frac{\sin(2x+2y)}{2x+2y}-1\right)\to 0$ which reduce to $\left(\frac{\sin(t)}{t}-1\right)\to 0$. $\endgroup$ – user Sep 23 '18 at 6:38

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