Algebra of limits for sequences

## Algebra of limits for sequences
 ---
 ## Algebra of limits
 #### Theorem
 If $\displaystyle\lim\_{n\to\infty}a_n$, $\displaystyle\lim\_{n\to\infty}b_n$ exist and $c\in\mathbb R$, then
 1. $\displaystyle\lim\_{n\to\infty}(a_n \pm b_n) = \left(\displaystyle\lim\_{n\to\infty}a_n\right) \pm \left(\displaystyle\lim\_{n\to\infty}b_n\right)$. 
 1. $\displaystyle\lim\_{n\to\infty}(a_n b_n) = \left(\displaystyle\lim\_{n\to\infty}a_n\right) \left(\displaystyle\lim\_{n\to\infty}b_n\right)$. 
 1. $\displaystyle\lim\_{n\to\infty}\left(\frac{a_n}{b_n}\right) = \frac{\displaystyle\lim\_{n\to\infty}a_n}{\displaystyle\lim\_{n\to\infty}b_n}$, provided $\displaystyle\lim\_{n\to\infty}b_n\neq0$. 
 1. $\displaystyle\lim\_{n\to\infty}(ca_n) = c\left(\displaystyle\lim\_{n\to\infty}a_n\right)$.

#### Proof
 See notes for 1. See problem set for 4. The hardest is 3; try it if you like.
 
 ---
 ## Algebra of limits: proof of product rule
 Theorem:
 If $\displaystyle\lim\_{n\to\infty}a_n$ and $\displaystyle\lim\_{n\to\infty}b_n$ exist, then
 $\displaystyle\lim\_{n\to\infty}(a_n b_n) = \left(\displaystyle\lim\_{n\to\infty}a_n\right) \left(\displaystyle\lim\_{n\to\infty}b_n\right)$.

#### Proof
 We know $(a_n)\_{n\in\mathbb N}$ is bounded, by $M$, say.
 We know $a_n \to \ell_a$ and $b_n\to\ell_b$ as $n\to\infty$, so
 \\[\begin{aligned}
 \&\forall \varepsilon_a>0 \\; \exists \\, k_a\in\mathbb N : n>k_a \implies \lvert a_n-\ell_a\rvert < \varepsilon_a, \\\\
 \&\forall \varepsilon_b>0 \\; \exists \\, k_b\in\mathbb N : n>k_b \implies \lvert b_n-\ell_b\rvert < \varepsilon_b.
 \end{aligned}\\]
 Fix $\varepsilon>0$.
 We calculate
 \\[\begin{aligned}
 \lvert a_nb_n - \ell_a\ell_b \rvert &= \lvert a_nb_n - a_n\ell_b + a_n\ell_b - \ell_a\ell_b \rvert \\\\
 &\leqslant \lvert a_n \rvert \lvert b_n-\ell_b \rvert + \lvert \ell_b \rvert \lvert a_n - \ell_a \rvert
 < \tfrac\varepsilon2 + \tfrac\varepsilon2,
 \end{aligned}\\]

provided $\lvert a_n-\ell_a \rvert < \frac\varepsilon {2\lvert \ell_b \rvert}$ and $\lvert b_n-\ell_b \rvert < \frac\varepsilon {2\lvert a_n \rvert}$.
 So let $\varepsilon_a = \frac\varepsilon {2\lvert \ell_b \rvert}$ and $\varepsilon_b = \frac\varepsilon {2M}$, and set $k=\max\\{k_a,k_b\\}$.
 Then, $\forall n>k$, $\lvert a_nb_n - \ell_a\ell_b \rvert<\frac\varepsilon2+\frac\varepsilon2=\varepsilon$. $\\;\square$
 
 ---
 ## Algebra of limits: proof of product rule
 Theorem:
 If $\displaystyle\lim\_{n\to\infty}a_n$ and $\displaystyle\lim\_{n\to\infty}b_n$ exist, then
 $\displaystyle\lim\_{n\to\infty}(a_n b_n) = \left(\displaystyle\lim\_{n\to\infty}a_n\right) \left(\displaystyle\lim\_{n\to\infty}b_n\right)$.

provided $\lvert a_n-\ell_a \rvert < \frac\varepsilon {2(1+\lvert \ell_b \rvert)}$ and $\lvert b_n-\ell_b \rvert < \frac\varepsilon {2(1+\lvert a_n \rvert)}$.
 So let $\varepsilon_a = \frac\varepsilon {2(1+\lvert \ell_b \rvert)}$ and $\varepsilon_b = \frac\varepsilon {2(1+M)}$, and set $k=\max\\{k_a,k_b\\}$.
 Then, $\forall n>k$, $\lvert a_nb_n - \ell_a\ell_b \rvert<\frac\varepsilon2+\frac\varepsilon2=\varepsilon$. $\\;\square$
 
 ---
 ## Algebra of limits: examples
 #### Problem
 Let $z_n = \frac3n + \frac{2+n-n^2}{17-2n+4n^2} + 5$.
 Calculate $\displaystyle \lim\_{n\to\infty} z_n$.
 
 ---
 ## Algebra of limits: examples
 #### Problem
 Let $z_n = \frac3n + \frac{2+n-n^2}{17-2n+4n^2} + 5$.
 Calculate $\displaystyle \lim\_{n\to\infty} z_n$.
 #### Attempted solution
 \\[
 \displaystyle \lim\_{n\to\infty} z_n
 = \lim\_{n\to\infty} \left(\frac3n\right) + \lim\_{n\to\infty} \left(\frac{2+n-n^2}{17-2n+4n^2}\right) + \lim\_{n\to\infty} \left(5\right)
 \\]

\\[
 = 3 \lim\_{n\to\infty} \left(\frac1n\right) + \frac{\lim\_{n\to\infty} \left(2+n-n^2\right)}{\lim\_{n\to\infty} \left(17-2n+4n^2\right)} + 5
 \\]
 
 ---
 ## Algebra of limits: examples
 #### Problem
 Let $z_n = \frac3n + \frac{2+n-n^2}{17-2n+4n^2} + 5$.
 Calculate $\displaystyle \lim\_{n\to\infty} z_n$.
 #### Attempted solution
 \\[
 \displaystyle \lim\_{n\to\infty} z_n
 = \lim\_{n\to\infty} \left(\frac3n\right) + \lim\_{n\to\infty} \left(\frac{2+n-n^2}{17-2n+4n^2}\right) + \lim\_{n\to\infty} \left(5\right)
 \\]

\\[
                = 3 \lim\_{n\to\infty} \left(\frac1n\right) + \frac{\lim\_{n\to\infty} \left(2+n-n^2\right)}{\lim\_{n\to\infty} \left(17-2n+4n^2\right)} + 5
            \\]

Error! Neither of the limits in the ratio exist. We misapplied rule 3.
 
 ---
 ## Algebra of limits: examples
 #### Problem
 Let $z_n = \frac3n + \frac{2+n-n^2}{17-2n+4n^2} + 5$.
 Calculate $\displaystyle \lim\_{n\to\infty} z_n$.
 #### Attempted solution
 \\[
 \displaystyle \lim\_{n\to\infty} z_n
 = \lim\_{n\to\infty} \left(\frac3n\right) + \lim\_{n\to\infty} \left(\frac{2+n-n^2}{17-2n+4n^2}\right) + \lim\_{n\to\infty} \left(5\right)
 \\]
 
 ---
 ## Algebra of limits: examples
 #### Problem
 Let $z_n = \frac3n + \frac{2+n-n^2}{17-2n+4n^2} + 5$.
 Calculate $\displaystyle \lim\_{n\to\infty} z_n$.
 #### Attempted solution
 \\[
 \displaystyle \lim\_{n\to\infty} z_n
 = \lim\_{n\to\infty} \left(\frac3n\right) + \lim\_{n\to\infty} \left(\frac{2+n-n^2}{17-2n+4n^2}\right) + \lim\_{n\to\infty} \left(5\right)
 \\]

\\[
 = 3 \lim\_{n\to\infty} \left(\frac1n\right) + \lim\_{n\to\infty} \left(\frac{\frac2{n^2}+\frac n{n^2}-\frac{n^2}{n^2}}{\frac{17}{n^2}-\frac{2n}{n^2}+\frac{4n^2}{n^2}}\right) + 5
 \\]
 
 ---
 ## Algebra of limits: examples
 #### Problem
 Let $z_n = \frac3n + \frac{2+n-n^2}{17-2n+4n^2} + 5$.
 Calculate $\displaystyle \lim\_{n\to\infty} z_n$.
 #### Attempted solution
 \\[
 \displaystyle \lim\_{n\to\infty} z_n
 = \lim\_{n\to\infty} \left(\frac3n\right) + \lim\_{n\to\infty} \left(\frac{2+n-n^2}{17-2n+4n^2}\right) + \lim\_{n\to\infty} \left(5\right)
 \\]

\\[
 = 3 \times 0 + \lim\_{n\to\infty} \left(\frac{\frac2{n^2}+\frac 1{n}-1}{\frac{17}{n^2}-\frac{2}{n}+4}\right) + 5
 \\]
 
 ---
 ## Algebra of limits: examples
 #### Attempted solution
 \\[
 \displaystyle \lim\_{n\to\infty} z_n
 = \lim\_{n\to\infty} \left(\frac3n\right) + \lim\_{n\to\infty} \left(\frac{2+n-n^2}{17-2n+4n^2}\right) + \lim\_{n\to\infty} \left(5\right)
 \\]

\\[
                = 3 \times 0 + \lim\_{n\to\infty} \left(\frac{\frac2{n^2}+\frac 1{n}-1}{\frac{17}{n^2}-\frac{2}{n}+4}\right) + 5
            \\]

\\[
 = 0 + \frac{\displaystyle\lim\_{n\to\infty} \left(\frac2{n^2}+\frac 1{n}-1\right)}{\displaystyle\lim\_{n\to\infty} \left(\frac{17}{n^2}-\frac{2}{n}+4\right)} + 5
 \\]
 
 ---
 ## Algebra of limits: examples
 #### Attempted solution
 \\[
 \displaystyle \lim\_{n\to\infty} z_n
 = \lim\_{n\to\infty} \left(\frac3n\right) + \lim\_{n\to\infty} \left(\frac{2+n-n^2}{17-2n+4n^2}\right) + \lim\_{n\to\infty} \left(5\right)
 \\]

\\[
                = \ldots \hspace{16em}
            \\]

\\[
 = 0 + \frac{\displaystyle\lim\_{n\to\infty} \left(\frac2{n^2}+\frac 1{n}-1\right)}{\displaystyle\lim\_{n\to\infty} \left(\frac{17}{n^2}-\frac{2}{n}+4\right)} + 5
 \\]
 
 ---
 ## Algebra of limits: examples
 #### Solution
 \\[
 \displaystyle \lim\_{n\to\infty} z_n
 = \lim\_{n\to\infty} \left(\frac3n\right) + \lim\_{n\to\infty} \left(\frac{2+n-n^2}{17-2n+4n^2}\right) + \lim\_{n\to\infty} \left(5\right)
 \\]

\\[
                = \ldots \hspace{16em}
            \\]

\\[
                = 0 + \frac{\displaystyle\lim\_{n\to\infty} \left(\frac2{n^2}+\frac 1{n}-1\right)}{\displaystyle\lim\_{n\to\infty} \left(\frac{17}{n^2}-\frac{2}{n}+4\right)} + 5
            \\]

\\[
 = 5 + \frac{\displaystyle\lim\_{n\to\infty} \left(\frac2{n^2}\right)+\lim\_{n\to\infty} \left(\frac 1{n}\right)-\lim\_{n\to\infty} \left(1\right)}{\displaystyle\lim\_{n\to\infty} \left(\frac{17}{n^2}\right)-\lim\_{n\to\infty} \left(\frac{2}{n}\right)+\lim\_{n\to\infty} \left(4\right)}
 = 5 + \frac{0+0-1}{0-0+4} = \frac{19}{4}.
 \\]
 
 ---
 ## Algebra of limits: tactic for application of quotient rule
 #### Theorem
 If $\displaystyle\lim\_{n\to\infty}a_n$ and $\displaystyle\lim\_{n\to\infty}b_n$ exist and $\displaystyle\lim\_{n\to\infty}b_n\neq0$, then
 $\displaystyle\lim\_{n\to\infty}\left(\frac{a_n}{b_n}\right) = \frac{\displaystyle\lim\_{n\to\infty}a_n}{\displaystyle\lim\_{n\to\infty}b_n}$.
 #### Observation
 To use this theorem, the lower limit must both exist and be nonzero.
 #### Tactic
 Divide the numerator and denominator by the fastest growing thing in the denominator.
 #### Example & Exercise: finish the calculation
 \\[
 \lim\_{n\to\infty} \frac{3n^3 - 4n+ \frac{n+1}{n+4}}{12 - 2n^3 + 4n^2}
 = \lim\_{n\to\infty} \frac{3 - \frac{4}{n^2}+ \frac{n+1}{n^3(n+4)}}{\frac{12}{n^3} - 2 + \frac4n}
 = \frac{\lim\_{n\to\infty} \left( 3 - \frac{4}{n^2}+ \frac{n+1}{n^3(n+4)}\right)}{\lim\_{n\to\infty} \left(\frac{12}{n^3} - 2 + \frac4n\right)}
 \\]
 
 ---
 ## Algebra of limits: tactic for application of quotient rule
 #### Example & Exercise: finish the calculation
 \\[
 \lim\_{n\to\infty} \frac{3n^3 - 4n+ \frac{n+1}{n+4}}{12 - 2n^3 + 4n^2}
 = \lim\_{n\to\infty} \frac{3 - \frac{4}{n^2}+ \frac{n+1}{n^3(n+4)}}{\frac{12}{n^3} - 2 + \frac4n}
 = \frac{\lim\_{n\to\infty} \left( 3 - \frac{4}{n^2}+ \frac{n+1}{n^3(n+4)}\right)}{\lim\_{n\to\infty} \left(\frac{12}{n^3} - 2 + \frac4n\right)}
 \\]

The denominator is $\displaystyle \lim\_{n\to\infty} \left(\frac{12}{n^3}\right) - \lim\_{n\to\infty} \left(2\right) + \lim\_{n\to\infty} \left(\frac4n\right) = 0 - 2 + 0 = -2 \neq 0$, so we were justified in using the quotient rule.

Numerator is $\displaystyle \lim\_{n\to\infty} \left(3\right) \\! - \\! \lim\_{n\to\infty} \left(\frac{4}{n^2}\right) \\! + \\! \lim\_{n\to\infty} \left(\frac{n+1}{n^3(n+4)}\right) = 3+\lim\_{n\to\infty} \left(\frac{\frac1{n^3}+\frac1{n^4}}{1+\frac4n}\right)$.

$= 3+\displaystyle\frac{\lim\_{n\to\infty} \left(\frac1{n^3}\right)+\lim\_{n\to\infty} \left(\frac1{n^4}\right)}{\lim\_{n\to\infty} \left(1\right)+\lim\_{n\to\infty} \left(\frac4n\right)} = 3+\frac{0+0}{1+0} = 3$.
 Again, we use that the limit of the denominator is nonzero to justify the second application of the quotient rule.
 
 ---
 ## Algebra of limits: more examples
 Let
 \\[\begin{aligned}
 a_n \&= \frac{700n^3 - \frac{2\sin(n)}{n} + \cos(n)n^2}{3n^2(-1)^n + 25n^3}, \\\\
 b_n \&= \frac{13n^2 - \frac\pi{n^2+7}}{n^3\sin(\pi n) + 7n^2 - 4}.
 \end{aligned}\\]
 Calculate $\displaystyle\lim\_{n\to\infty} a_n$ and $\displaystyle\lim\_{n\to\infty} b_n$.

Calculate also $\displaystyle\lim\_{n\to\infty} \left(5a_nb_n\right)$, $\displaystyle\lim\_{n\to\infty} \left(a_n^2(b_n+4)\right)$ and $\displaystyle\lim\_{n\to\infty} \left(\frac{a_n-1}{7b_n}\right)$.