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I am trying to prove the following limits using the delta-epsilon method. Can you help me out?

$$ \lim_{(x,y)\to(a,b)}(x+y) = a+b$$

user92821
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  • You want to prove that $$\forall (a,b)\in \Bbb R^2\forall \delta>0 \exists \varepsilon >0\forall(x,y)\in \Bbb R^2\left(|(x,y)-(a,b)|<\varepsilon\implies |(x+y)-(a-b)|<\delta\right).$$

    Let $(a,b)\in \Bbb R^2$ and $\delta >0$ be taken arbitrarily.

    I only checked in my head, but $\varepsilon =\dfrac \delta 2$ seems to work.

     – Git Gud Oct 10 '13 at 20:43

2 Answers2

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Hint: For $\epsilon > 0$, choose $\delta = \epsilon / 2$. Then show that if $|x-a| < \delta$ and $|y-b| < \delta$, then $|(x+y) - (a+b)| < \epsilon$. This should suffice for whatever choice of metric you want to use on ${\mathbb R}^2$, getting the point across without particulars about choice of metric etc. Intuitively it just shows that as $(x,y)$ gets close to $(a,b)$, you get that $x+y$ gets close to $a+b$.

user2566092
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  • This proof implicitly uses the $\infty$-metric... – AlexR Oct 10 '13 at 20:49
  • I agree, but the question is whether they're even considering explicit different metrics in limit problems. My guess is that the $\infty$-metric is the canonical one for simple problems like this. It also is the metric that appeals the most to intuition, i.e. choosing one $\delta$ for all variables. – user2566092 Oct 10 '13 at 20:52
  • It sure does and the equivalency of norms assures us the correct result independent of the norm, but essentially this is no different than the proof I stated, only a tad bit less exact. Referring to @GitGud, we should probably find a way around this (i.e. using the euclidean norm). – AlexR Oct 10 '13 at 20:54
  • Ideally a well-posed limit problem should either state a norm or state that we can assume all norms are equivalent. I agree your answer subsumes mine, I was just trying to side-step the quagmire of different norms by appealing to the most intuitive one, and making a side-comment that it should be intuitively obvious that this $\infty$-norm is sufficient for any other one. – user2566092 Oct 10 '13 at 20:58
  • I agree on this point, the norm should have been mentioned in the problem or the lecture. To clarify the equivalence of at least the two most common norms, $\Vert\cdot\Vert_\infty$ and $\Vert\cdot\Vert_2$, I added a note to my answer. – AlexR Oct 10 '13 at 21:01
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Hint:
Since on $\mathbb R^n$ all norms are equivalent, chose the easiest: $\Vert\cdot\Vert_\infty$. This will leave you with two 1-D. $\epsilon$-$\delta$ arguments where $$\epsilon = (\epsilon_1, \epsilon_2)$$ and you want $$|x+y-a-b| = |(x-a)+(y-b)| \leq |x-a| + |y-b| < \epsilon_1 + \epsilon_2 = 2\Vert\epsilon\Vert_\infty$$ Get the gist? I can provide more detail if needed.

Now all that remains to prove is $$\lim_{x\to a} x = a$$ or equivalently (by exchanging symbols) $$\lim_{y\to b} y = b$$ For the choice of $\delta$, $\delta_\epsilon := \frac{\epsilon}{2}$ suits your needs.

Proof
Using equivalency of norms on $\mathbb R^n$, we have to prove $$\forall\ \epsilon > 0 \quad \exists\ \delta > 0$$ such that $$|x+y - (a+b)| < \epsilon \qquad \forall\ (x,y) \in \mathbb R^2 | \Vert (x,y) - (a,b) \Vert_\infty < \delta$$ Now for $\delta := \frac{\epsilon}{2}$ we have $$|x+y - (a+b)| \leq |x-a| + |y-b| < \delta + \delta = 2\delta = \epsilon$$ q.e.d.

Note
Using the euclidean norm $\Vert\cdot\Vert_2$ we can chose $\delta = \frac{\epsilon}{4}$, since $$\Vert (x,y) \Vert_2^2 = x^2 + y^2 \leq x^2 + 2|x||y| + y^2 = (|x| + |y|)^2 \leq (2\Vert(x,y)\Vert_\infty)^2$$ (actually we give away a bit here, but that's not important for the task at hand)

AlexR
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  • Thank you! I don't really understand what you chose for delta. What is this dot? I am sorry, I am still learning the basics! Also, What do you mean that this leaves us with two 1-D. ε-δ arguments? – user92821 Oct 10 '13 at 20:40
  • $\cdot$ is used as a placeholder, and with 1-D. I mean one-dimensional. For the concrete stuff, see the edit – AlexR Oct 10 '13 at 20:41
  • @user92821 Any questions remaining? If not I suggest you mark the question as answered. – AlexR Oct 10 '13 at 21:00
  • You can still use $\Vert\cdot\Vert_\infty$, but I guess a different choice of $\delta$ is needed ;-). Essentially that means yes, you can. // Besides you have shown in the original problem, that you can take the limits per summand and add up afterwards. – AlexR Oct 10 '13 at 21:03
  • Okay! Thanks so much AlexR. – user92821 Oct 10 '13 at 21:06