real analysis

a) I assume the text means that $\lim_{n \rightarrow \infty} c_n = M$ (instead of $b_n$). It is False, because the limit of $b_n$ might not exist, take for example $a_n = -1$, $c_n = 1$, and $b_n = \sin(n)$. If the limit of $b_n$ exists (say it's $R$), however, then it lies in the interval $[L, M]$, because for any $\epsilon$ there exists $n$ lager enough so that $b_n \geq a_n \geq L - \epsilon$, so $R \geq L$, and similarly for $R \leq M$.

b) False again. Take $a_{2n+1} = 1$ and $a_{2n} = 2$ for all $n$. Then $\limsup_{n \rightarrow \infty} a_n = 2$, but $\liminf_{n \rightarrow \infty} b_n = -1$.