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Question: I kinda forgot my multivariable limits already...

Is it legit to put a condition under a limit eg $\lim_{\substack{(x,y) \to (0,0) }} \frac{x}{x^2+y^2}$ doesn't exist because

$$\lim_{\substack{(x,y) \to (0,0) \\ y=0 }} \frac{x}{x^2+y^2}$$

$$= \lim_{\substack{(x,y) \to (0,0) \\ y=0 }} \frac{x}{x^2}$$

$$= \lim_{\substack{(x,y) \to (0,0)}} \frac{x}{x^2}$$

$$= \lim_{\substack{x \to 0}} \frac{x}{x^2}$$

$$= \lim_{\substack{x \to 0}} \frac{1}{x}$$

doesn't exist

?

Explanation: The thing is in single variable real: it's legit to say $\lim_{x \to 0^+} \frac 1 x$ because there's a precise definition for this where instead of $- \delta < x - 0 < \delta$ you say $0 < x - 0 < \delta$ or something. Hence, here it's legit to write $\lim_{\substack{x \to 0 \\ x > 0}} \frac 1 x := \lim_{x \to 0^+} \frac 1 x$.

I would like to know this is similarly legit in complex/real multivariable, if it is. I think if this isn't legit, then I would like to know the proper way to argue. Perhaps say $\frac{x}{x^2+y^2} = \frac{1}{x}$ for $y=0$ and thus... (wait let's change to $x=0$) let's say $\frac{x}{x^2+y^2} = 0$ for $x=0$ and thus $\frac{x}{x^2+y^2} \to 0$ as $y \to 0$ along $x=0$...but I don't know how to make this precise either.

(Context: I'm actually asking with complex limits in mind, but I guess it will be equivalent to real multivariable limits anyway.)

Guess:

Ah wait writing all this I think I figured it out.

(convert to answer)

BCLC
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2 Answers2

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If a notation hasn't been defined, it's technically not legit to use it.

On the other hand, some extensions of notations have so obvious meaning that they hardly need an explicit definition. IMHO your notation belongs to this class. Therefore I consider it legit.

md2perpe
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  • 'If a notation hasn't been defined, it's technically not legit to use it.' --> God bless you, md2perpe. thank you. – BCLC Oct 06 '21 at 18:29
  • ok but basically adding a condition under a limit is just adding a condition on the $z$ or $(x,y)$ or $x$ on top of the $0 < |\cdot|< \delta$ thing as described in my guess answer? – BCLC Oct 06 '21 at 18:30
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    The obvious interpretation of $\lim_{\substack{x\to a \ P(x) }} f(x) = L$ is $$\forall \epsilon>0 , \exists \delta>0 ,\forall x : (|x-a|<\delta \land P(x) \implies |f(x)-L| < \epsilon).$$ – md2perpe Oct 06 '21 at 18:40
  • md2perpe ok...thanks – BCLC Oct 07 '21 at 12:27
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Guess:

The generalisation of changing $-\delta < x - 0 < \delta$ to $0 < x - 0 < \delta$ is that we take the intersection of $- \delta < x - 0 < \delta$ with $x > 0$. Therefore,

the original definition of

$$\lim_{\substack{x \to a}} f(x) = L$$

as

$$\forall \varepsilon > 0, \exists \delta > 0 \ \text{s.t.} \ |f(x)-L| < \varepsilon \ \text{whenever} \ 0<|x-a|<\delta \ (\text{and} \ x \in Domain(f))$$

extends to

$$\lim_{\substack{x \to a \\ \text{condition on x}}} f(x) = L$$ defined as

$$\forall \varepsilon > 0, \exists \delta > 0 \ \text{s.t.} \ |f(x)-L| < \varepsilon \ \text{whenever}$$

$$\ 0<|x-a|<\delta \ (\text{and} \ x \in Domain(f)) \ \text{and} \ \text{condition on x}$$

Therefore in multivariable or even complex the legit-ness (legitimacy) of conditions is like...

$$\lim_{\substack{(x,y) \to (0,0) \\ x=0 }} \frac{x}{x^2+y^2}$$ defined as

$$\forall \varepsilon > 0, \exists \delta > 0 \ \text{s.t.} \ |\frac{x}{x^2+y^2}-0| < \varepsilon \ \text{whenever}$$

$$\ 0<\sqrt{(x-0)^2+(y-0)^2}<\delta \ (\text{and} \ (x,y) \in Domain(f)) \ \text{and} \ \{x=0\}$$

(where $f: Domain(f) \to \mathbb R, f(x,y)=\frac{x}{x^2+y^2}$)

Edit: To further increase legitimacy:

$$\text{condition on x}$$

is merely a fancy way of saying

$$x \in A$$

for some subset $A \subseteq \mathbb C$ or $\mathbb R^2$

BCLC
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