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How do I solve the below limit

$$\tag{1} \lim_{(x,y)\to (0,0)} \frac{\sin(x+y)}{x+y}?$$

My approach:

I used Polar Coordinates $x = r \cos(\theta)$ and $y = r \sin(\theta)$

so (1) => $$\tag{2} \lim_{r\to 0} \frac{\sin(r (\cos(\theta) + \sin(\theta))}{r (\cos(\theta) + \sin(\theta))}.$$

And then first solution:

I set $w = r (\cos(\theta) + \sin(\theta)$ so when $r\to 0 $ and $w\to 0$

(2) $\Rightarrow \lim_{w\to 0} \frac{sin(w)}{w}= 1$.

Second solution: L'Hospital's rule:

\begin{align} (2) \Rightarrow \lim_{r\to 0} \frac{(\sin(r (\cos(\theta) + \sin(\theta)))'}{(r (\cos(\theta) + \sin(\theta)))'}& = \lim_{r\to 0} \frac{\cos(r (\cos(\theta) + \sin(\theta))*(\cos(\theta) + \sin(\theta))}{\cos(\theta) + \sin(\theta)}\\ & = \lim_{r\to 0} {\cos(r (\cos(\theta) + \sin(\theta))} = \cos(0) = 1. \end{align}

Is my approaches correct? If not can you provide a correct solution?

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  • $\begingroup$ It could also be proved by the Taylor series of $sin$. $\endgroup$ – user723846 Aug 20 '20 at 16:50
  • $\begingroup$ @Gio So my approaches are correct? $\endgroup$ – Andreas Aug 20 '20 at 17:18
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You can use that $z(x,y)=x+y$ and $f(t)=\frac{\sin t}{t}$ are continuous functions and their superposition is also continuous.

Rudin W. - Principles of mathematical analysis-(1976) page 86. Theorem 4.7

Suppose $X,Y,Z$ are metric spaces, $E \subset X$, $f$ maps $E$ into $Y$, $g$ maps the range of $f,f(E)$, into $Z$, and $h$ is the mapping of $E$ into $Z$ defined by $h(x)=g(f(x)), x \in E$. If $f$ is continuous in point $p \in E$ and $g$ is continuous at the point $f(p)$, then $h$ is continuous at $p$.

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  • $\begingroup$ Are my approaches correct? $\endgroup$ – Andreas Aug 20 '20 at 16:58
  • $\begingroup$ @Andreas. Firstly let me note that $x = r \cos(\theta), y = r \sin(\theta)$ is known under name of polar coordinates for Cartesian $(x,y)$. I understand your way so you consider function $(r, \theta) \to (x,y)$. But then you wrote $y = r (\cos(\theta) + \sin(\theta))$, so again use letter "$y$". Do I correctly understand that these are 2 different "$y$"? Dependent on your answer I'll continue. $\endgroup$ – zkutch Aug 20 '20 at 17:51
  • $\begingroup$ I edit my post. Yes its different 'y' i rename the variable to w. Thanks for your corrections. $\endgroup$ – Andreas Aug 20 '20 at 18:17
  • $\begingroup$ @Andreas. Ok. So you have $(x,y) \to (r, \theta) \to w \to z=\frac{\sin w}{w}$ - now to apply limit you need extra details accordingly polar coordinates and if you conduct everything well, then it should work. But why you take extra level using polar coordinates, what you gain? In answer suggested by me I am making same, but with only 2 functions. Now about L'Hospital: I have doubt it is correct, because you show, that limit exist with fixed $\theta$ i.e. with respect to each line. But there are examples, that it is not enough for double limit. $\endgroup$ – zkutch Aug 20 '20 at 20:14
  • $\begingroup$ So the first solution seems correct from what you say? I used polar coordinate because i did not find other solution. "In answer suggested by me I am making same, but with only 2 functions." I do not understand your solution can you explain it further? $\endgroup$ – Andreas Aug 20 '20 at 21:15
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Let's start from the definition of multivariate limits: to say $\lim_{(x,\,y)\to(0,\,0)}\frac{\sin(x+y)}{x+y}=L$ is equivalent to$$\forall\epsilon>0\exists\delta>0\forall(x,\,y)\left(0<\sqrt{x^2+y^2}<\delta\to\left|\frac{\sin(x+y)}{x+y}-L\right|<\epsilon\right).$$We can prove this for $L=1$ using$$\forall\epsilon>0\exists\delta^\prime>0\forall(x,\,y)\left(0<|x+y|<\delta^\prime\to\left|\frac{\sin(x+y)}{x+y}-1\right|<\epsilon\right).$$We just need to choose $\delta$ in terms of $\delta^\prime$ so $\sqrt{x^2+y^2}<\delta\to|x+y|<\delta^\prime$. It suffices to take $\delta=\delta^\prime/2$ (proof is an exercise); in fact, it suffices to take $\delta=\delta^\prime/\sqrt{2}$ (proof is a slightly harder exercise).

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  • $\begingroup$ Are my approaches correct? $\endgroup$ – Andreas Aug 20 '20 at 16:58
  • $\begingroup$ @Andreas Your approaches assume $(x,\,y)\to(0,\,0)$ is equivalent, regardless of path, to $r\to0^+$ at fixed $\theta$. This is provable with techniques such as mine. $\endgroup$ – J.G. Aug 20 '20 at 17:07
  • $\begingroup$ What do you meam with 'my approaches assume (x,y)→(0,0) is equivalent' ? $\endgroup$ – Andreas Aug 20 '20 at 17:19
  • $\begingroup$ @Andreas Equivalent regardless of path to etc. $\endgroup$ – J.G. Aug 20 '20 at 17:54

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