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## Show limit using limit theorems carefully

Solution: By Theorem 9.4, we have
$$\lim a_n^3=(\lim a_n)^3=a^3,\quad \lim b_n^2=(\lim b_n)^2=b^2$$ By Theorem 9.2, we have
$$\lim 4a_n=4\lim a_n=4a.$$ Therefore, we have, by Theorem 9.4,
$$\lim (a_n^3+4a_n)=\lim a_n^3+\lim 4a_n=a^3+4a$$ and
$$\lim (b_n^2+1)=\lim b_n^2 +1=b^2+1.$$ Clearly, we have $\lim (b_n^2+1)>0$, hence we can use Theorem 9.6 and obtain
$$\lim s_n=\frac{\lim(a_n^3+4a_n)}{\lim(b_n^2+1)}=\frac{a^3+4a}{b^2+1}.$$ We are done.