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I'm currently making the transition from single variable calculus to multivariable calculus, and the epsilon-delta proofs seem as daunting as ever. I'm currently stuck on this one: $$\lim\limits_{(x,y) \to (1,2)} \ x^2 +2y = 5$$

It seems really simple but I am not being able to find a relation between the epsilon and the delta. Can you help me?

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$\forall\varepsilon>0$ there exist $\delta>0$ such that if $0<\sqrt{(x-1)^2+(y-2)^2}<\delta$ then $$|x^2 +2y - 5|=|(x-1)(x+1)+2(y-2)|\leq|x+1||x-1|+2|y-2|<(\delta+2)\delta+2\delta$$ since $|x-1|<\sqrt{(x-1)^2+(y-2)^2}<\delta$ thus $2-\delta<x+1<2+\delta$.

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I'll give you some hints: first of all, it's easier for calculations to change variables and consider the (equivalent) limit:

$$\lim\limits_{(x,y) \to (0,0)} \ (x+1)^2 +2(y+2) = 5 $$

We shall prove that given any $\epsilon > 0 $ there exists a $\delta > 0 $ such that $|f(x,y) - 5 | < \epsilon $ whenever $0 < \sqrt{x^2+y^2} < \delta $. We have:

$$|f(x,y) - 5 |= |x^2 + 2(x+y) | \le |x^2|+2|x+y| $$

Now, note that for every $x,y$, $|x+y| \le 2\sqrt{x^2+y^2}$, can you choose a $\delta=\delta(\epsilon)$ and conclude?

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  • $\begingroup$ how do I deal with the module of x^2? Do I make x = 1 and proceed? $\endgroup$ – BSD Apr 11 '17 at 17:43
  • $\begingroup$ No, you absolutely can't say that! Remember that now $(x,y) \rightarrow (0,0) $, but you can't even set $x=0$...we are working "near" zero, not at zero. You must also bound $x^2$ with some quantity that you can control, that is a quantity function of $x^2+y^2$ $\endgroup$ – GaC Apr 11 '17 at 18:11
  • $\begingroup$ Yes, when I said 'set' I meant 'bound' x (sorry but these specific terms in english are tricky for me). Let's set ${\sqrt (x^2 + y^2)} = 1$. That leaves me with $x^2 >1- y^2$. But how can I replace it now? I am not being able to connect this last part to the delta. $\endgroup$ – BSD Apr 11 '17 at 18:32

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