Best way to evaluate $\lim_{n \rightarrow \infty} |\frac{(3(n+1)+4)(4^{n+1}+5)(5^n+3)}{(5^{n+1}+3)(3n+4)(4^n+5)}|$ Best way to evaluate $\lim_{n \rightarrow \infty} |\frac{(3(n+1)+4)(4^{n+1}+5)(5^n+3)}{(5^{n+1}+3)(3n+4)(4^n+5)}|$
Can you all show me some different ways to evaluate this limit? I was thinking of multiplying the numerator and denominator by something.. maybe $\frac{1}{5^{n+1}}$... I don't know. I'm sure there are a lot of ways to evalute this and you guys will show me a lot of slick awesome ways!!
As you may have guessed, i'm in the midst of employing the ratio test for a series.
 A: We need to factor out the leading terms to obtain
$$\frac{(3(n+1)+4)(4^{n+1}+5)(5^n+3)}{(5^{n+1}+3)(3n+4)(4^n+5)}
=\frac{(n+1)\cdot 4^{n+1} \cdot 5^n}{5^{n+1} \cdot n \cdot 4^n}\frac{\left(3+\frac4{n+1}\right)\left(1+\frac5{4^{n+1}}\right)\left(1+\frac3{5^n}\right)}{\left(1+\frac3{5^{n+1}}\right)\left(3+\frac4n\right)\left(1+\frac 5 {4^n}\right)}$$
with
$$\frac{(n+1)\cdot 4^{n+1} \cdot 5^n}{5^{n+1} \cdot n \cdot 4^n}=\frac{4}{5}\frac{n+1}n \to \frac45$$
and
$$\frac{\left(3+\frac4{n+1}\right)\left(1+\frac5{4^{n+1}}\right)\left(1+\frac3{5^n}\right)}{\left(1+\frac3{5^{n+1}}\right)\left(3+\frac4n\right)\left(1+\frac 5 {4^n}\right)} \to1$$
A: It is worth noting that in general, for real numbers $a$ and $b$ you have
\begin{eqnarray*}
\lim_{n\to\infty}\frac{a^{n+1}+b}{a^n+b}
&=&\lim_{n\to\infty}\frac{a^{n+1}+ab-ab+b}{a^n+b}\\
&=&\lim_{n\to\infty}\frac{a(a^n+b)+(1-a)b}{a^n+b}\\
&=&\lim_{n\to\infty}\left(a+\frac{(1-a)b}{a^n+b}\right)\\
&=&a+(1-a)b\lim_{n\to\infty}\frac{1}{a^n+b}.
\end{eqnarray*}
So if $a>1$ the latter limit equals $0$, and so we get
$$\lim_{n\to\infty}\frac{a^{n+1}+b}{a^n+b}=a.$$
Of course it is a basic result on limits that for polynomials $f$ and $g$ with leading coefficients $u$ and $v$
$$\lim_{n\to\infty}\frac{f(n)}{g(n)}=\frac{u}{v},$$
so in particular $\lim_{n\to\infty}\frac{f(n+1)}{f(n)}=1$. So we can break your example into three simpler limits:
$$L_1:=\lim_{n\to\infty}\frac{3(n+1)+4}{3n+4}=1,\qquad L_2:=\lim_{n\to\infty}\frac{4^{n+1}+5}{4^n+5}=4,\qquad L_3:=\lim_{n\to\infty}\frac{5^{n+1}+3}{5^n+3}=5,$$
and then your limit equals
$$\lim_{n \rightarrow \infty} |\frac{(3(n+1)+4)(4^{n+1}+5)(5^n+3)}{(3n+4)(4^n+5)(5^{n+1}+3)}|=L_1\cdot L_2\cdot L_3^{-1}=\frac{4}{5}.$$
A: @user's answer is the common way when you have not learned equivalents yet.
Later whenever $\lim\dfrac{f(n)}{g(n)}=1$ noted $f(n)\sim g(n)$
you will go on and replace all the terms of a product by their simpler equivalents.
$\begin{cases}3(n+1)+4&\sim 3n\\4^{n+1}+5&\sim 4^{n+1}\\5^n+3&\sim 5^n\\5^{n+1}+3&\sim 5^{n+1}\\3n+4&\sim 3n\\4^n+5&\sim 4^n\end{cases}\quad$ and get $f(n)\sim\dfrac{3n\ 4^{n+1}\ 5^n}{5^{n+1}\ 3n\ 4^n}=\dfrac 45$
This is a tool which allows to make a writing shortcut of "factoring the dominant term" method, that @user has shown.
A: Here is an alternative solution. The limit at hand is of the form $\lim_n\frac{a_{n+1}}{a_n}$, which if exists, then it is equal to $\lim_n\sqrt[n]{a_n}$ by a well known result.
So, for
$$ a_n=\frac{(3n+1)(4^n+5)}{5^n+3}$$
we have
$$ 
\sqrt[n]{a_n} = \frac{4}{5}\frac{(3n+1)^{1/n}\big(1+\frac{5}{4^n}\big)^{1/n}}{\big(1+\frac{3}{5^n}\big)^{1/n}}$$

*

*Since $(3n)^{1/n}\leq (3n+1)^{1/n}<(4n)^{1/n}$, we have $(3n+1)^{1/n}\xrightarrow{n\rightarrow\infty}1$.

*Since $1\leq (1+4^{-n}5)^{1/n}\leq 2^{1/n}$ (for $\geq2$), we have $(1+4^{-n}5)^{1/n}\xrightarrow{n\rightarrow\infty}1$.

*Similarly, $(1+5^{-n}3)^{1/n}\xrightarrow{n\rightarrow\infty}1$.

Putting things together, $\lim_n\sqrt[n]{a_n}=\lim_n\frac{a_{n+1}}{a_n}=\frac{4}{5}$.
