# limit trough path

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 : $$\lim_{x\rightarrow 0} f(x) =L$$

• We can just apply it to the path $\gamma(t)=t-b$, unless I'm missing something ? – Suzet Feb 13 '20 at 14:07
• my mistake i wrote it wrong – Tair Galili Feb 13 '20 at 14:15

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 $$t_n. Such a path $$\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.
• 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) – Tair Galili Feb 13 '20 at 14:50
• @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 ! – Suzet Feb 13 '20 at 14:58