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I am just trying to find a formal proof for this statement:

Let $f:\mathbb R^n \backslash \ \{0\} \rightarrow \Bbb R$ be a function, and assume that there exits $L\in \Bbb R$ that for every path $$\gamma: (a,b)\rightarrow \Bbb R^n\backslash\{0\} $$ that satisfies $$\lim_{t\rightarrow b}\gamma(t) = 0 $$ the limit $$\lim_{t\rightarrow b} f(\gamma(t)) = L $$ show that : $\lim_{x\rightarrow 0} f(x) =L$

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    $\begingroup$ We can just apply it to the path $\gamma(t)=t-b$, unless I'm missing something ? $\endgroup$
    – Suzet
    Feb 13, 2020 at 14:07
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    $\begingroup$ my mistake i wrote it wrong $\endgroup$ Feb 13, 2020 at 14:15

1 Answer 1

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With the edited question, let us assume towards a contradiction that $\lim_{x\rightarrow 0} f(x) \not = L$ (the limit either doesn't exist or if it does, it isn't $L$).

There exists some positive $\epsilon$ such that for any $n\geq 0$, there is some $x_n\in B^{\star}(0,\frac{1}{n})$ (open ball centered at $0$ with radius $\frac{1}{n}$ and $0$ removed) such that $||f(x_n)-L||\geq \epsilon$.

For $n\geq 0$, let $t_n:=-2^{-n}$ so that $t_0=-1$ and $t_n$ goes to $0$ as $n$ goes to infinity.
Because $B^{\star}(0,\frac{1}{n})$ is path-connected, for every $n\geq 0$ there exists some path $\gamma_n:[t_{n},t_{n+1}] \rightarrow B^{\star}(0,\frac{1}{n})$ with $\gamma(t_{n})=x_{n}$ and $\gamma(t_{n+1})=x_{n+1}$.
I define the path $\gamma : (-1,0)\rightarrow \Bbb R^n\backslash\{0\}$ by letting $\gamma(t)=\gamma_n(t)$ where $n\geq 0$ is the unique nonnegative integer such that $t_n<t\leq t_{n+1}$. Such a path $\gamma$ is well defined, it glues all the $\gamma_n$ together.

Now, by construction I have $\gamma(t)\in B^{\star}(0,\frac{1}{n})$ for every $t\in (t_n,0)$, which implies that $\lim_{t\rightarrow 0}\gamma(t) = 0$. Meanwhile, $f(\gamma(t_n))=f(x_n)$ for every positive $n$, which implies $$||f(\gamma(t_n))-L||\geq \epsilon$$ whence $f(\gamma(t))$ can not converge to $L$ as $t$ goes to $0$, this is the desired contradiction.

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  • $\begingroup$ thanks a lot! I thought that the proof is going to be shorter but this one is really nice though. (just in the first part, $n$ cant be zero... but it really doesnt matter for the convergence) $\endgroup$ Feb 13, 2020 at 14:50
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    $\begingroup$ @tairgalili It can be made shorter if we skip over the details, constructing the path $\gamma$ is in the end just more difficult to write than to think of ! $\endgroup$
    – Suzet
    Feb 13, 2020 at 14:58

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