Prove / disprove if $\lim((a_{n})^2+ (b_{n})^2)=0$ then $\lim(a_{n})=0$ and $\lim(b_{n})=0$

Question

Prove / disprove: Let $a_{n}$ and $b_{n}$ be real sequences, if $\lim((a_{n})^2+ (b_{n})^2)=0$ then $\lim(a_{n})=0$ and $\lim(b_{n})=0$

Answer 1

Hint:   $0 \le a_n^2 \le a_n^2+b_n^2\,$, so $\,a_n^2 \to 0\,$ by the squeeze theorem.

Answer 2

Suppose $a_n$ does not go to 0. Then there exists an $\epsilon>0$ such that $|a_n|>\epsilon$ for infinitely many $n$. This implies $a_n^2>\epsilon^2>0$ for infinitely many $n$. Since $a_n^2+b_n^2\geq a_n^2$, this implies $\lim (a_n^2+b_n^2)\geq \epsilon^2>0$, which implies that the original limit could not possibly be 0.

Answer 3

Since $$|a_n^2+b_n^2|<\epsilon^2$$so is $$|a_n^2|\le |a_n^2+b_n^2|<\epsilon^2$$