Question

Prove if |x_sub(n) - L| <= b_sub(n) and lim b_sub(n)=0, then lim x_sub(n)=L

Answer 1

Question 1. Prove if $|x_{n} - L| \leq b_{n}$ and $\lim b_{n} = 0$, then $\lim x_{n} = L$.

Solution. Note that $|x_{n} - L| \leq b_{n}$ implies $b_{n} \geq 0$. Since $\lim b_{n} = 0$, for each $\varepsilon > 0$ there is $N \in \mathbb{N}$ such that $0 \leq b_{n} < \varepsilon$ for all $n > N$. Then $|x_{n} - L| \leq b_{n} < \varepsilon$ for all $n > N$. By definition this means that $\lim x_{n} = L$.