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I want to prove that $\lim_{(x,y)\to (0,0)} \frac{\sin(x+y)}{x+y} = 1$

Is it sufficient to say that if $u = x+y$ then as $(x,y) \rightarrow (0,0)$ then $u \rightarrow 0$ and then evaluate $\lim_{u\to 0} \frac{\sin u}{u}$ which is 1 by L'Hôpital's rule?

I feel that this is somehow "cheating" and I would really like to prove it in a more rigorous way, i.e., by using an $\epsilon$ - $\delta$ argument.

I have tried for a long time to use the one variable version and prove that if $0 \lt \sqrt{x^2 + y^2} \lt \delta$ then somehow $0 \lt |x+y| \lt \delta$ and it would follow from the one variable case that $| \frac{\sin(x+y)}{x+y} - 1| \lt \epsilon$. However, because $|x+y| = 0$ when $y = -x$ I am not getting anywhere with this.

Any input on this would be helpful :)

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I'm sorry. I don't know LaTeX and I didn't realize that I needed the $'s. Edited it now. – Skull_Kid Jan 31 '13 at 18:40
You are right, the points $(x,-x)$ for $x\neq 0$ create a problem in that the function is not defined everywhere on any neighborhood $\{(x,y)\;;\;0<\sqrt{x^2+y^2}<\delta\}$. So the limit does not exist. – 1015 Jan 31 '13 at 18:41
@Skull_kid:its not cheating.theorems are tools that help me to solve problem easily.but you want solve this question by using epsilon- delta – Maisam Hedyelloo Jan 31 '13 at 18:41
It is definition dependent. For a starker example, consider the limit of $\frac{xy}{xy}$. Some would say the limit is $1$. Some would worry about the fact that the function is not defined at $(10^{-7},0)$. – André Nicolas Jan 31 '13 at 18:43
@julien : The fact that the function is not defined everywhere in any neighbourhood of (0,0) would not imply that the limit does not exist, now will it? According to the definition of a limit; our function must be defined on a subset A of R^n, then if x0 is either in A or is a boundary point of A the limit exists if and only if 0 <|| x - x0 || < delta blablabla. In this case (0,0) is a boundary point of our A. A is just R^2 \ (the line y = - x). That is R^2 with the mentioned line removed. Besides, I'm pretty sure the limit is 1. Just look at the plot of the function. – Skull_Kid Jan 31 '13 at 19:05
up vote 5 down vote accepted

As @julien points out, the function $\frac{\sin(x+y)}{x+y}$ is not defined in a neighbourhood of the origin, so the limit does not exist. But there is a natural extension of the function which is defined on a neighbourhood of the origin, and for which the limit exists and equals 1. (There are lots of extensions, but this one seems most natural to me.) Define $f:\mathbb{R}^2\to\mathbb{R}$ by $$f(x,y)=\left\{ \begin{array}{cl} \frac{\sin(x+y)}{x+y} & x+y\neq0; \\ 1 & x+y=0.\end{array}\right.$$ Then $\lim_{(x,y)\to(0,0)}f(x,y)=1$. An $\epsilon-\delta$ proof relies on the one variable result $$\lim_{u\to 0}\frac{\sin u}{u}=1.$$ That is, given $\epsilon>0$, there exists $\delta>0$ such that $$0<|u|<\delta \Rightarrow \left|\frac{\sin u}{u}-1\right|<\epsilon.$$ It is helpful to notice that this can be rephrased as follows: given $\epsilon>0$, there exists $\delta>0$ such that $$0<|u|<2\delta \Rightarrow \left|\frac{\sin u}{u}-1\right|<\epsilon. \qquad (*)$$

We want to prove that given $\epsilon>0$, there exists a positive $\delta$ such that $$0<\sqrt{x^2+y^2}<\delta \Rightarrow |f(x,y)-1|<\epsilon.$$

So let $\epsilon>0$. Choose $\delta$ so that $(*)$ holds true, and let $0<\sqrt{x^2+y^2}<\delta$. Then $$|x+y|\leq|x|+|y|\leq\sqrt{|x|^2+|y|^2}+\sqrt{|x|^2+|y|^2}=2\sqrt{x^2+y^2}<2\delta.$$ Then by (*) (for $x+y\neq0$) and by the definition of $f$ (for $x+y=0$), we have $|f(x,y)-1|<\epsilon$ as required.

The same argument works if the function is defined as $g:\mathbb{R}^2\setminus\{(x,y):x+y=0\}\to\mathbb{R}$ with $g(x,y)=\frac{\sin(x+y)}{x+y}$. With a slight amendment, the same result also holds for $$f(x,y)=\left\{ \begin{array}{cl} \frac{\sin(x+y)}{x+y} & x+y\neq0; \\ h(x) & x+y=0,\end{array}\right.$$ where $h$ is any function defined in a neighbourhood of $x=0$ which has $\lim_{x\to0}h(x)=1$.

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Thanks a lot. Great reply, that also goes to the rest of you :) I was actually VERY close to this. I just struggled to find an inequality of the form $|x+y| \lt k\cdot\sqrt{x^2 + y^2}$, for some constant k. Because I needed both to be smaller than $\delta$ and some function of $\delta$ – Skull_Kid Jan 31 '13 at 23:15

Presumably, when we talk about $\lim_{(x,y)\to(0,0)}\frac{\sin(x+y)}{x+y}$, the domain of $\frac{\sin(x+y)}{x+y}$ is assumed to be $\mathbb{R}^2\setminus\{(x,y):x=y\}$, otherwise, as you and the others have pointed out, the limit does not exist. Now, given that the issue has been settled, it's perfectly OK to evaluate the limit as $\lim_{u\to0}\frac{\sin u}{u}$. Let $u=x+y$ and $v=x-y$. Recall what is meant by $\lim_{(x,y)\to(0,0)}\frac{\sin(x+y)}{x+y}=L$:

(a) For any $\epsilon>0$, there exists $\delta>0$ such that $\|(u,v)\|<\delta \Rightarrow \left|\frac{\sin(u)}{u}-1\right|<\epsilon$.

And what is meant by $\lim_{u\to0}\frac{\sin(u)}{u}=L$ is this:

(b) For any $\epsilon>0$, there exists $\delta>0$ such that $|u|<\delta \Rightarrow \left|\frac{\sin(u)}{u}-1\right|<\epsilon$.

So, suppose you have shown (b). Then, given any $\|(u,v)\|<\delta$, you will have $|u|<\delta$. Hence by (b), you get (a).

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Use the fact that $|x+y|\leq \sqrt{2}\sqrt{x^2+y^2}$.

If $\epsilon>0$ and $\delta>0$ are such that $|\dfrac{\sin u}{u}-1|<\epsilon$ for $|u|<\delta$ then
for $(x,y)$ with $\|(x,y)\|=\sqrt{x^2+y^2}<\dfrac{\delta}{\sqrt{2}}\Longrightarrow$ ...

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How do you prove that $|x+y| \le \sqrt{2}\cdot\sqrt{x^2 + y^2}$ ? And is this true for all $(x,y) \in \R2$ ? – Skull_Kid Jan 31 '13 at 22:49
That should be $ \mathbb{R}^2$ – Skull_Kid Jan 31 '13 at 22:59
@Skull_Kid: Is true $\forall (x,y)\in\mathbb R^2$ and follows from C-S inequality. To prove it use that $(x+y)^2\leq 2(x^2+y^2), \ \forall (x,y)\in\mathbb R^2$. The last one follows from $0\leq (x-y)^2=2(x^2+y^2)-(x+y)^2$. – P.. Feb 1 '13 at 5:40

$$\lim_{(x,y)\to (0,0)} \frac{\sin(x+y)}{x+y} $$ $$1) let\ y=0 \ Then \lim_{(x,0)\to (0,0)} \frac{\sin(x)}{x} = 1= L_1$$ $$2)let\ y=0 \ Then \lim_{(0,y)\to (0,0)} \frac{\sin(y)}{y} = 1=L_2$$ $$3)\ let\ y=x \ Then \lim_{(x,x)\to (0,0)} \frac{\sin(2x)}{2x} = 1=L_3$$


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There are many other ways of approaching (0,0) than the three that you mentioned. – minimalrho Jan 31 '13 at 19:24

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