Show the limit is zero
Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 9 Exercise 9.7 Solution: We have seen that $0\le s_n<\sqrt{\dfrac{2}{n-1}}$. By Sequeeze-Theorem/Exercise 8.5, it suffices to show that $\lim…
Proof of Squeeze Lemma/Theorem
Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 8 Exercise 8.5 Solution: Part a Let $\epsilon>0$. Since $\lim a_n=s$, there exists $N_1>0$ such that $$ |a_n-s|<\epsilon\Longrightarrow -\epsilon…