Which solution is correct: $\lim_{x \to \infty} {\frac{x^2+2x+3}{2x^2+x+5}}^\frac{3x-2}{3x+2}$? I was solving this question $\lim_{x \to \infty} {\frac{x^2+2x+3}{2x^2+x+5}}^\frac{3x-2}{3x+2}$.
Now, this is solved using $\lim_{x \to a} [1+f(x)]^\frac{1}{g(x)}=e^{\lim_{x \to a} \frac{f(x)}{g(x)}}$.
However, there seem to be two methods of approaching this problem and both yield different answers. In the first approach we multiply by $\frac{1}{2}$ outside the limit and by $2$ inside the limit and add and subtract $1$ to get in the form of the formula. After solving we get the value of the limit as $e$. In the second approach we add and subtract $1$ inside the limit thus getting it in this form and then evaluating the limit we get $e^{-\frac{1}{2}}$.
Now, the book prefers the first approach and so I am assuming my second approach is wrong, but I can't find for the life of me where it is wrong.

Here is how your second approach should work:
$$\frac{x^2+2x+3}{2x^2+x+5} = 1 + \left(\frac{x^2+2x+3}{2x^2+x+5}-1\right)$$$$ = 1 + \underbrace{\frac{-x^2+x-2}{2x^2+x+5}}_{\stackrel{x\to\infty}{\longrightarrow}-\frac 12}$$
So, you could not use "your rule with $e$", because it would require this term to tend to zero.
So, you actually get
$$\left( 1 +\frac{-x^2+x-2}{2x^2+x+5}\right)^{\frac{3x-2}{3x+2}}\stackrel{x\to\infty}{\longrightarrow}\left(1-\frac 12\right)^1 = \frac 12$$
