Proof that a lipschitz continuous function has a limit

Question

Let f:ℝ{0} => ℝ be a lipschitz continous function and a a sequence with lim a = 0 and image(a) ⊂ ℝ{0}. I have to proof that ∃limit(f(an)n∈ℕ). And I have to decide if the statement is also correct for uniformly continuous functions.

My thought was that because of limits of sequences for continuous functions it directly follow that there has to be a limit for f in both cases, but I dont know if this is correct.

Thanks in advance for any kind of help

Answer 1

Recall that $f$ is continuous at $x$ if and only if $f(x_n)\to f(x)$ whenever $x_n\in x$. In this question you can't conclude the result from continuity because $0$ is not in the domain of the function. One can easily construct a continuous function $f:\mathbb{R}\setminus\{0\}\to\mathbb{R}$ such that $\lim_{x\to 0}f(x)$ does not exist. For example $f(x)=\sin\frac{1}{x}$.

Indeed you will have to use Lipschitz condition. Since $f$ is Lipschitz, there exists $C>0$ such that $$|f(x)-f(y)|\leq C|x-y|,\quad\forall x,y\in\Bbb{R}\setminus\{0\}.$$ In particular $$|f(a_n)-f(a_m)|\leq C|a_n-a_m|,\quad\forall m,n\in\Bbb{N}.$$ Note that $(a_k)$ is convergent so it's a Cauchy sequence. By definition of Cauchy sequence, for every $\epsilon>0$, there exists $N_0\in\Bbb{N}$ such that $$|a_n-a_m|N_0.$$ This means $$|f(a_n)-f(a_m)|<\epsilon,\quad\forall n,m>N_0.$$ Thus $(f(x_k))_{k=1}^\infty$ is a Cauchy sequence. Hopefully this may also give you some idea for the uniform continuous case.