Problems from Operations and limits with sequences

Calculate the limit of the following sequences;

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

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

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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 $$\dfrac{7}{4}$$. Hide solution and development Calculate the limit of the sum, subtraction, multiplication and division, when possible, of the following pairs of sequences; a) $$a_n=\dfrac{n^3-3}{n-6}$$ and $$b_n=\dfrac{3n^4-7}{3n^2+n-1}$$ b) $$a_n=2+3^n$$ and $$b_n=\dfrac{1}{5^n}$$ See development and solution Development: a) The limit of the two sequences is infinite and for the defined arithmetical rules we see that the limit of the sum and that of the product are infinite. To calculate the limit of the subtraction and of the quotient we must solve the indeterminate form. For the limit of the subtraction; $$\lim_{n \to \infty}{\dfrac{n^3-3}{n-6}-\dfrac{3n^4-7}{3n^2+n-1}}=\lim_{n \to \infty}{\Big(\dfrac{(n^3-3)(3n^2+n-1)-(3n^4-7)(n-6)}{(n-6)(3n^2+n-1)}\Big)}=$$$ $$=\lim_{n \to \infty}{\Big(\dfrac{19n^4-n^3-9n^2+4n-39}{3n^3-5n^2-7n+6}\Big)}=\infty$$$For the limit of the quotient; $$\lim_{n \to \infty}{\dfrac{n^3-3}{n-6}/\dfrac{3n^4-7}{3n^2+n-1}}=\lim_{n \to \infty}{\Big(\dfrac{(n^3-3)(3n^2+n-1)}{(n-6)(3n^4-7)}\Big)}=$$$ $$=\lim_{n \to \infty}{\Big(\dfrac{3n^5+n^4-n^3-9n^2-3n+3}{3n^5-18n^4-7n+42}\Big)}=1$$$b) The $$a_n=2+3^n$$ has limit infinity and the $$b_n=\dfrac{1}{5^n}$$ has limit $$0$$. Following the arithmetical rules, the limit of the sum and subtraction are infinity. To calculate the limit of the product; $$\lim_{n \to \infty}{(2+3^n)\cdot\dfrac{1}{5^n}}=\lim_{n \to \infty}{\dfrac{2}{5^n}}+\lim_{n \to \infty}{\dfrac{3^n}{5^n}}=0+0=0$$$

To calculate the limit of the quotient; $$\lim_{n \to \infty}{(2+3^n)/\dfrac{1}{5^n}}=\lim_{n \to \infty}{2\cdot5^n}+\lim_{n \to \infty}{3^n\cdot5^n}=\infty+\infty=\infty$$\$

Solution:

a) The limit of the sum, subtraction and multiplication is infinite. The limit of the quotient is $$1$$.

b) The limit of the sum, subtraction and quotient is infinite. The limit of the product is $$0$$.

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