I've been working on this question for about 2 hours and haven't gotten far (I'm not the best at this area) I'm not really looking for a full answer just someone to put me in the right direction as I'd understand similair examples more in the future, unless you wanna write up a full answer no ones stopping you haha 🙂 Anyway, here's the full question and my working up until now:
Question:
Consider a function $f:\mathbb{R}\rightarrow\mathbb{R}$ and a point $a\in\mathbb{R}$. Assume that $\lim_{x\to a}f(x)=l$ for some $l\in\mathbb{R}$. Prove that $\lim_{n\to\infty}f(x_n)=l$ for every sequence $(x_n)_{n=1}^{\infty} \subset (\mathbb{R}\backslash(a))$ such that $\lim_{n\to\infty}x_n=a$
Working:
Disclaimer: My working may be a bit jumbled because I usually take notes and right down ideas, then when I get a solution I re-write it and I haven't actually got a solution yet (evidently), so sorry if it's a bit confusing! 🙂
$\lim_{x\to a}f(x)=l \implies |x-a|<\epsilon$ and $|f(x)-l|<\delta$
Assume $\lim_{n\to\infty}f(x)=L$
Fix $\epsilon>0$ then $\exists\delta>0 : \forall x : 0<|x-a|<\delta$
Then $|f(x)-L|<\delta$
Fix $\epsilon$>0 for $(x_n)^{\infty}_{n=1}\implies\exists N : n>N\implies|x_n-a|<\epsilon$
Any help would greatly be appreciated! 🙂
Best Answer
Let $\epsilon>0$. Since $$\lim_{x\to a}f(x)=l$$ we can find $\delta>0$ such that $$0<|x-a|<\delta\implies |f(x)-l|<\epsilon.$$ Since $$\lim_{n\to\infty}x_n=a$$ then corresponding to $\delta>0$, we can find $N\in\Bbb N$ such that for all $n\geq N$ we get $|x_n-a|<\delta$. We know that $x_n\neq a$ for all $n\in\Bbb N$. Thus, for all $n\geq N$, we get $0<|x_n-a|<\delta$ and hence $|f(x_n)-l|<\epsilon$ and we are done.