If $x_n >0$ is a sequence of positive real numbers and $x_n\to x$ as $n \to \infty$, prove that $x \ge 0$.
Suppose $x<0$. Since $x_n \to x$, then for all $\epsilon$ there exists $N$ such that $n \ge N \implies |x_n-x|< \epsilon$.
I know we must have infinitely many terms of the sequence close to $x$, but does this say there must exist infinitely many negative terms in the sequence? If so, how?
Best Answer
Assume that $x<0$, then there exists some $N$ such that if $n\geq N$, then $|x_{n}-x|<-x/2$, then $x_{n}=x_{n}-x+x\leq|x_{n}-x|+x<-x/2+x=x/2<0$.