limsup of a functon and sequences

Question

I would like to know if this statement is true: if $\limsup_{x \rightarrow x_0} f(x)=\lambda$ then exists a sequence $\{x_n\}$ that converges to $x_0$ such that $\lim_{n \rightarrow \infty } f(x_n)= \lambda$. Intuitively i think this is true, but i would like to understand why it is true or false.

Answer 1

Yes, it is true.

Let $E$ be the domain of $f$. By definition, $\limsup_{x\to x_0}f(x)=\lambda$ means that$$\lim_{\delta\to0}\bigl(\sup\{f(x)\,|\,x\in E\cap(x_0-\delta,x_0+\delta)\}\bigr)=\lambda.$$Take $\varepsilon>0$. Then, by the previous equality,$$\sup\left\{f(x)\,\middle|\,x\in E\cap\left(x_0-\frac1n,x_0+\frac1n\right)\right\}\in[\lambda,\lambda+\varepsilon)$$for some $n\in\mathbb N$. So, there is some $x_n\in\left(x_0-\frac1n,x_0+\frac1n\right)$ such that $x_n\in[\lambda,\lambda+\varepsilon)$ and this implies that $\bigl|\lambda-f(x_n)\bigr|<\varepsilon$. Since $(\forall n\in\mathbb{N}):|x_0-x_n|<\frac1n$, $\lim_{n\to\infty}x_n=x_0$.