# Using combination rules!!! Help please!

• January 7th 2010, 05:07 AM
reeha
for(a), it is false. Here is a counterexaple: $a_n=n+1, b_n=n,\frac{a_n}{b_n}\to 1$, but $a_n-b_n=1\to 1$.
for(b), It is ture! $\left|a_n-b_n\right|=\left|b_n\right|\cdot\left|\frac{a_n}{b _n}-1\right|\to 0$ since $b_n$ is bounded.