Algebra of Limits for Sequences

$$\def \C {\mathbb{C}}\def \R {\mathbb{R}}\def \N {\mathbb{N}}\def \Z {\mathbb{Z}}\def \Q {\mathbb{Q}}\def\defn#1{{\bf{#1}}}$$
Theorem Algebra of Limits
Suppose that $(a_n)$ and $(b_n)$ are two convergent sequences with $a_n\to a$ and $b_n\to b$. Then the following are true:

(i) Scalar rule: $ka_n\to ka$ for all $k\in \R$. Proof

(ii) Sum Rule: $a_n +b_n \to a+b$. Proof

(iii) Product Rule: $a_n b_n \to ab$. Proof

(iv) Quotient Rule: If $b_n\neq 0$ for all $n\in \N $ and $b\neq 0$, then $a_n / b_n \to a/b$. Proof

(v) Square Root Rule: If $a_n\geq 0$ for all $n$, then $\sqrt{a_n} \to \sqrt{a} $. Proof


Show that $ a_n = \frac{3n^2}{n^2-5} \to 3$.

The naive way to do this would be to say $a_n=3n^2$ and $b_n=n^2-5$ and attempt to apply the quotient rule. Unfortunately, then as $n$ gets larger $a_n$ and $b_n$ get larger and don’t converge to a limit. Instead we pull another trick. We divide top and bottom by $n^2$ as follows:

We have
\lim_{n\to \infty } a_n&= \lim_{n\to \infty }\dfrac{3n^2/n^2}{(n^2-5)/n^2} \\
&=\lim_{n\to \infty } \dfrac{3}{1-\dfrac{5}{n^2} } \\
&= \dfrac{\displaystyle \lim_{n\to \infty }3}{\displaystyle \lim_{n\to \infty } \left( 1-\dfrac{5}{n^2} \right)} ,{\text{ by the quotient rule (if the limit is non-zero — it will be!)}},\\
&=\dfrac{3}{\displaystyle \lim_{n\to \infty }1 – \lim_{n\to \infty } \dfrac{1}{n^2} }, {\text{ by the sum rule as the limits exist}},\\ \\
&=\dfrac{3}{1} = 3.

$$c_n = \dfrac{3n^2+5n}{4n^2-5} .$$
Show that $c_n\to \dfrac{3}{4} $.

We have
&=& \dfrac{3n^2+5n}{4n^2-5} \\
&=& \dfrac{(3n^2/n^2)+(5n/n^2)}{(4n^2/n^2)-(5/n^2)} \\
&=& \dfrac{3+5/n}{4-(5/n^2)} \\
&\to & \dfrac{3+0}{4+0} {\text{ by the sum and quotient rules}}, \\

Find the limit, if it exists, of $\dfrac{2n^5}{(3n^3+2)(2n+1)^2} $ as $n\to \infty $.

We have,
\lim_{n\to \infty} \dfrac{2n^5}{(3n^3+2)(2n+1)^2}
&= \lim_{n\to \infty} \dfrac{2n^5/n^5}{(1/n^3)(3n^3+2)(1/n^2)(2n+1)^2} \\
&=\lim_{n\to \infty} \dfrac{2}{(3+2/n^3)(2+1/n)^2} \\
&= \dfrac{2}{(3+0)(2+0)^2} \\
&=\dfrac{1}{6} .

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