Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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, $|x_i - a_i| \le \parallel x - a\parallel \lt \delta, thus \parallel f(x_i) - v\parallel < \epsilon$

share|improve this question

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.