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](https://www.math24.net/number-e).

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 $e$. 

[![enter image description here](https://i.stack.imgur.com/s4uGx.png)](https://i.stack.imgur.com/s4uGx.png)