I'm asking this because I'm having a hard time to prove the case of $x \lt 0$ for this question: Suppose that $x \in R, X_{n} \geq 0, and X_{n} \rightarrow x \: as \: n \rightarrow \: \infty$, Prove that $\sqrt{X_{n}} \: \rightarrow \: \sqrt{x} \: as \: n \: \rightarrow \: \infty$, since the assumption of $x \in R$ means x can be negative. But please don't directly answer my homework for me. I'm hoping to find if such assumption is correct first then continue my work for it. Thanks!
Are there sequences of positive real numbers converge to negative real number
convergence-divergencelimitsreal-analysissequences-and-series
Best Answer
That cannot happen.
Suppose $x_n \to y$, where $y<0$ and $x_n >0$ for all $n$. Consider the interval $(y-\epsilon, y+\epsilon)$ around $y$, where $\epsilon < \frac{|y|}{2}$. In this neighborhood, can any $x_n$ be there?