Question statement: Let $f : \mathbb{R}^n\to\mathbb{R}$ be a function. If for any sequence $\vec{r_n}\to\vec{a}$, the sequence $f(\vec{r_n})$ converges to $L$, then prove that $\lim_{\vec{r} \to \vec{a}} f(\vec{r}) = L$
My try : The only way I could think of was by proving the contrapositive. Thus now, $\lim_{\vec{r} \to \vec{a}} f(\vec{r})$ does not exist. This implies, there exists $\epsilon>0$, s.t. for every $\delta>0$ whenever $0<||\vec{r}-\vec{a}||<\delta$ , $|f(\vec{r})-L|\geq\epsilon$ for some $L\in\mathbb{R}$ . I now use this to show that whenever $0<||\vec{r_n}-\vec{a}||<\delta\ \ \forall n\geq N(\delta)$ , $|f(\vec{r_n})-L|\geq\epsilon$ . Thus for all sequences $\{\vec{r_n}\}$ converges to $\vec{a}$ , the sequence $\{f(\vec{r_n})\}$ does not converge.
Is this line of thought correct ? Also, even if it is correct/wrong, is there any other approach to solving the above mentioned problem instead of the contrapositive method ?
You're correct in your approach, but your quantifiers need some remediation.
Suppose we can find $\epsilon>0$ such that for any $\delta>0$ there exists $r_0\in\mathbb{R}^n-\{a\}$ with $\|r_0-a\|<\delta$ and $|f(r_0)-L|\geq \epsilon$.
This means, for any $k\in \mathbb{N}$, there is $r_k\in \mathbb{R}^n-\{a\}$ such that $\|r_k-a\|<\frac{1}{k}$ but $|f(r_k)-L|\geq \epsilon$.
Clearly $\{r_k\}_{k\geq 1}$ converges to $a$ but $\{f(r_k)\}_{k\geq 0}$ does not converge to $L$, which is your contradiction.