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If we want to show that the limit of function of two variables doesn't exist , it is sufficient two paths which have different limits . Intuitively , it seems correct but how we can prove it with the using of definition of limit ?

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  • $\begingroup$ Have you tried to prove it? Where are you having difficulty? $\endgroup$
    – saulspatz
    Mar 1, 2019 at 15:54
  • $\begingroup$ @saulspatz Yes I tried , I don't know how a path is related to definition of limit . $\endgroup$
    – S.H.W
    Mar 1, 2019 at 15:55
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    $\begingroup$ The distance between two distinct points (the limits, in this case) is positive, say $d$. Consider the definition of a limit with $\varepsilon=d/4$. $\endgroup$
    – Clayton
    Mar 1, 2019 at 15:58
  • $\begingroup$ A path is a continuous function. $\endgroup$
    – saulspatz
    Mar 1, 2019 at 15:58
  • $\begingroup$ @Clayton Can you explain more please ? $\endgroup$
    – S.H.W
    Mar 1, 2019 at 16:02

1 Answer 1

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Denote by $D$ the domain of your function, and pick $x_0 \in D$.

If you want the classic $\varepsilon - \delta$ definition of continuity of $f$ at $x_0$, here it is: $$\forall \varepsilon >0, \ \ \ \exists \delta >0 : \ \ \ \forall x \in D, \ \ \ |x-x_0|< \delta \Rightarrow |f(x)-f(x_0)| < \varepsilon \ \ \ \ \ \ (*)$$ Let's write the condition: "for some path the limit is $L$" $$\gamma_1:[0,1] \to D \ \ \mathrm{continuous}, \ \ \ \gamma_1(1)=x_0, \ \ \ \lim_{t \to 1}f( \gamma_1 (t))=L$$

A similar condition for some second path, with limit $L' \neq L$:

$$\gamma_2:[0,1] \to D \ \ \mathrm{continuous}, \ \ \ \gamma_2(1)=x_0, \ \ \ \lim_{t \to 1}f( \gamma_2 (t))=L'$$

Now, pick $\varepsilon = \frac{1}{9}|L-L'|$, and $\delta$ accordingly to $(*)$. By continuity of $\gamma_1$ and $\gamma_2$ at $t=1$, there esists $\delta_2$ such that $$\forall t \in [0,1], \ \ |1-t|< \delta_2 \Rightarrow (|\gamma_1(t)- x_0| < \delta \ \ \ \ \wedge \ \ \ |\gamma_2(t)-x_0| < \delta)$$ Thus, by $(*)$ you have $$|f(\gamma_1(t)) - f(x_0)| < \varepsilon \ \ \ \wedge \ \ \ |f(\gamma_2(t)) - f(x_0)| < \varepsilon$$

Moreover, there exists $\delta_3 >0$ such that $$\forall t \in [0,1], \ \ |1-t|< \delta_3 \Rightarrow (|f(\gamma_1(t))- L| < \varepsilon \ \ \ \ \wedge \ \ \ |f(\gamma_2(t))-L'| < \varepsilon)$$ Finally, take any $t \in [0,1]$ with $t < \delta_2, \delta_3$. Then $$|f( \gamma_1(t)) - f( \gamma_2(t))| = |(f( \gamma_1(t)) -f(x_0)) - ( f( \gamma_2(t))- f(x_0))| \le 2 \varepsilon= \frac{2}{9}|L-L'|$$ on the other hand $$|f( \gamma_1(t)) - f( \gamma_2(t))|= |(f( \gamma_1(t))-L) - (f( \gamma_2(t))-L') + (L-L')|\ge |L-L'|- 2 \varepsilon = \\ = \frac{7}{9}|L-L'|$$ which is a contradiction.

Now, you can see that writing all of this using $\varepsilon - \delta$ becomes very heavy and tedious: this is the reason why nobody wanted to answer you in these terms.

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  • $\begingroup$ Why $\gamma_1$ and $\gamma_2$ have $[0 , 1]$ domain ? $\endgroup$
    – S.H.W
    Mar 1, 2019 at 16:34
  • $\begingroup$ They are paths. Do you know what a path is? $\endgroup$
    – Crostul
    Mar 1, 2019 at 16:39
  • $\begingroup$ Not exactly , I only encountered some examples like $y = mx$ or $x = y^2$ . $\endgroup$
    – S.H.W
    Mar 1, 2019 at 16:45
  • $\begingroup$ Well, for example, $y=mx$ is a formula for the path $\gamma(x)=(x,mx)$, where $x \to 0$. Now, if you call $t=1-x$, you have a continuous function of $t \to 1$ with $\gamma (1)=x_0$. $\endgroup$
    – Crostul
    Mar 1, 2019 at 16:48
  • $\begingroup$ Sorry , I'm really confused . Can you give a precise definition of path please ? $\endgroup$
    – S.H.W
    Mar 1, 2019 at 16:54

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