Calculate the limit of the following sequences;

a) $$a_n=\dfrac{(-1)^n\cdot(2n+3)}{9n^3-11n+3}$$

b) $$a_n=\dfrac{(-1)^nn^2+7n^3+21}{4n^3-8}$$

### Development:

a) The sequence is given by the product of the sequences $$a_n=(-1)^n$$ and $$b_n=\dfrac{(2n+3)}{9n^3-11n+3}$$.

The first one is bounded and the second one has limit $$0$$. Applying the presented result we obtain that the product has limit $$0$$.

b) In this case we cannot apply the result used to the previous paragraph. We can say that the limit of the sum is the sum of limits. Namely, $$$\lim_{n \to \infty}{a_n}=\lim_{n \to \infty}{\dfrac{(-1)^nn^2}{4n^3-8}}+\lim_{n \to \infty}{\dfrac{7n^3+21}{4n^3-8}}$$$

Now we can apply the result commented upon regarding the first limit. Since it corresponds to the product of the sequences $$\{(-1)^n\}_{n\in\mathbb{N}}$$ and $$\Big\{\dfrac{7n^3+21}{4n^3-8} \Big\}_{n\in\mathbb{N}}$$.

Then the first limit of the sum is $$0$$. The second limit is $$\dfrac{7}{4}$$.

This way we obtain the desired result $$\lim_{n \to \infty}{a_n}= \dfrac{7}{4}$$.

### Solution:

a) The limit is $$0$$.

b) The limit is $$\dfrac{7}{4}$$.