Finding a limit of a sequence that requires simplification using $e$ and properties of $e$. The problem is the following: 

Study the limit of the sequence: $\left(\dfrac{3n+4}{3n-2}\right)^{2n+1}$ as  $n \to \infty$

Below is a picture of the solution of the problem I received. However, I find it hard to understand what happens in line 5 where we put our equation to the $e$ to the power of something. I know the identities for $e$ when taken as a limit of $n \to 0$ or infinity as explained here.
I tried to rewrite the equation as was done at other problems from the link above but I get a 'too complicated equation' where instead of getting $(1- \frac1x)^x$ or $(1+u)^u$ I get negative roots when I make a substitution and can't get the equation to the corresponding form. 
Going back to the solution below, I cannot understand how they took it to the power of $e$ and especially why is there $a-1$ (where we have $f(n)-1$) In other words, how did they transform: 
$$ \left(\frac{3+\frac{4}{n}}{3-\frac{2}{n}}\right)^{2n+1} $$
to something to the power of $e$. 

 A: They want to use $\lim_{t\to 0}\left(1+t\right)^{ \frac 1t}=e$.
So, starting with


*

*$\lim_{x\to a}f(x) = 1 \Leftrightarrow \lim_{x\to a}(\underbrace{f(x)-1}_{=t(x)})=0$, they put

*$f(x) = 1 + (f(x)-1)$ 

*$\Rightarrow \left(1+(f(x)-1)\right)^{\frac{1}{f(x)-1}} \stackrel{t(x)=f(x)-1}{=} \left(1+t(x)\right)^{ \frac 1{t(x)}}\stackrel{t(x) \to 0}{\longrightarrow}e$
In your case, the calculation can be done a bit easier as follows:
$$ \left(\frac{3n+4}{3n-2}\right)^{2n+1} =\left(\frac{3n-2+6}{3n-2}\right)^{2n+1} = \left(1+\frac{6}{3n-2}\right)^{2n+1}$$
$$\Rightarrow  \left( \underbrace{\left(1+\frac{6}{3n-2}\right)^{\frac{3n-2}{6}} }_{\stackrel{n\to\infty}{\rightarrow}e}\right)^{\underbrace{\frac{6}{3n-2}\cdot(2n+1)}_{\stackrel{n\to\infty}{\rightarrow}\frac{12}{3}=4}}\stackrel{n\to\infty}{\rightarrow}e^4$$
A: Consider $$a_n=\left(\frac{3n+4}{3n-2}\right)^{2n+1}\implies \log(a_n)=(2n+1)\log\left(\frac{3n+4}{3n-2}\right)=(2n+1)\log\left(1+\frac{6}{3n-2}\right) $$ Since $n$ is large, using equivalents,
$$\log\left(1+\frac{6}{3n-2}\right)\sim\frac{6}{3n-2}\implies \log(a_n)\sim\frac{6(2n+1)}{3n-2}\sim 4\implies a_n \sim e^4$$
You could meke it more general for
$$x_n=\left(\frac{a n+b}{c n+d}\right)^{e n+f}$$ and show that
$$\log(x_n)=e \log \left(\frac{a}{c}\right)n+\left(\frac{b e}{a}-\frac{d e}{c}+f \log
   \left(\frac{a}{c}\right)\right)$$ So, a finite limit requires $c=a$ making the limit to be
$$\exp\left(\frac{e (b-d)}{a} \right)$$
A: When $L=\lim_{x \rightarrow a} f(x)^{g(x)} \rightarrow (1)^\infty$ (indeterminate), this means $\lim_{x \rightarrow a} f(x) =1+\epsilon$ and $\lim_{x \rightarrow a} \rightarrow  \infty$. We know that $X=\exp[\ln X].$ So we can write:
$$L=\exp[\lim_{x\rightarrow a} f(x)^{g(x)}=\exp[\lim_{x \rightarrow a} g(x) \ln(f(x))]=\exp[\lim_{x\rightarrow a} g(x) \ln(1+\epsilon)].$$
Next use $\ln(1+\epsilon) \approx \epsilon$, and next $\epsilon \approx f(x)-1$.
Finally, ewe get $$L=\exp[\lim_{x \rightarrow a} g(x) (f(x)-1)].$$
This is how you have solved this limit in your notes.
A: write the expression as $\Bigg(\{1+\frac{1}{\frac{3n-2}{6}}\}^{\frac{3n-2}{6}}\Bigg)^{\frac{12n+6}{3n-2}}$
now see that,  $\lim_{n\to \infty} \Bigg(\{1+\frac{1}{\frac{3n-2}{6}}\}^{\frac{3n-2}{6}}\Bigg)= e$ and also that $\lim_{n\to\infty} {\frac{12n+6}{3n-2}}=4$, so the final answer is $e^4$.
