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Consider $f:S \subseteq \mathbb{R}^n \to \mathbb{R}^m$ and two points; $a$ a limit point of $S$ and $v$ in $\mathbb{R}^m$. Show that $\displaystyle \lim_{x\to a}f(x) = v$ iff for each sequence ($x_k$) in $S \backslash \{a\}$ with $\displaystyle \lim_{x\to \infty}x_k = a$, we have $\displaystyle \lim_{k\to \infty}f(x_k) = v$ .

Honestly, I understand it intuitively but am stumped as to how i would prove it. Any direction would be appreciated.

attempt: Forward direction, Suppose $\lim_{x\to a}f(x) = v$, then for all $\epsilon > o$ there exists a $\delta$ >0 such that $\parallel x - a\parallel < \delta$ implies $\parallel f(x) - v\parallel < \epsilon$. since $\parallel x - a\parallel = (\sum_{i = 0}^{\infty}(x_i - a_i)^2)^{1/2} \ge |x_i - a_i|$, therefore, $|x_i - a_i| \le \parallel x - a\parallel \lt \delta, thus \parallel f(x_i) - v\parallel < \epsilon$

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up vote 1 down vote accepted

Suppose $\lim_{x\to a}f(x)=v$. Then for every $\varepsilon>0$ we have a $\delta>0$ such that $$\Vert x-a\Vert<\delta_1\Longrightarrow\Vert f(x)-v\Vert<\varepsilon$$ by definition of limit. Now, suppose $x_k\to a$ as $k\to\infty$. By definition, this means that for every $\delta_2>0$ there exists an $N>0$ such that $$\Vert x_n-a\Vert<\delta_2$$ for $n>N$. Let $\delta=\min\{\delta_1,\delta_2\}$. Let $N_0\in\Bbb N$ be large enough so that $$\Vert x_{N_0}-a\Vert<\delta\leq\delta_1\Longrightarrow\Vert f(x_{N_0})-v\Vert<\varepsilon.$$ Since $\varepsilon>0$ was arbitrary and $\{x_k\}_{k=1}^\infty$ was an arbitrary sequence, we're done.

The converse uses similar ideas.

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