Getting basic step for calculating the limit of e I have the following limit:
$$
\lim_{x\to \infty}{\bigg( \frac{x^2 + 3x - 1}{x^2 + 3} \bigg)  }^\frac{x -2}{2}
$$
We had no explanations for calculating such limits i looked over few poor textbook examples and i understand the result of this limit will be $ e^X $ where X will be what i get from expanding the limits $ \frac {x-2}{2} $.  
And that first step is to get the following form:
$$
\lim_{x\to\infty}{ \bigg ( 1 + \frac{1}{x+1} \bigg )^{x+1} } = e
$$
Are there any simpler steps to solving such a limit ?
 A: $$\lim_{x\to \infty}{\left( \frac{x^2 + 3x - 1}{x^2 + 3} \right)  }^\frac{x -2}{2}=\lim_{x\to \infty}{\left( \frac{x^2 + 3-3+3x - 1}{x^2 + 3} \right)  }^\frac{x -2}{2}=\lim_{x\to \infty}{\left( \frac{x^2 + 3+3x - 4}{x^2 + 3} \right)  }^\frac{x -2}{2}=\lim_{x\to \infty}{\left( 1+\frac{3x - 4}{x^2 + 3} \right)  }^\frac{x -2}{2}=\lim_{x\to \infty}{\left( 1+\frac{3x - 4}{x^2 + 3} \right)  }^{\frac{x^2+3}{3x-4}\cdot \frac{3x-4}{x^2+3}\cdot\frac{x -2}{2}}\overset{def}{=}L$$
Since exists $\lim\limits_{x\to \infty}{\left( \dfrac{3x-4}{x^2+3}\cdot\dfrac{x -2}{2} \right)  }=\dfrac{3}{2}$, then $L=e^\frac{3}{2}.$
A: Another approach:
If $\lim\limits_{x\to{+\infty}} f(x)^{g(x)}$ is $1^{+\infty}$, which is an indeterminate form, then: $$\lim_{x\to{+\infty}} f(x)^{g(x)}=e^{\lim\limits_{x\to +\infty}\big(f(x)-1\big)g(x)}$$
A: You're right that you want to use the fact that
$$\lim_{x\to\infty} f(x) = \lim_{x\to\infty} e^{F(x)},$$ 
where $F = \log f$. In this case, using L'hopital's rule,
\begin{align*}\lim_{x\to\infty} \frac{x-2}{2} \log \left(\frac{x^2+3x-1}{x^2+3}\right) &= \lim_{x\to\infty} \frac{ \log \left(\frac{x^2+3x-1}{x^2+3}\right) }{\frac{2}{x-2}}\\
&= \lim_{x\to\infty} \frac{\frac{-3x^2+8x+9}{(x^2+3)(x^2+3x-1)}}{\frac{-2}{(x-2)^2}}\\
&= \lim_{x\to\infty} \frac{(x-2)^2(-3x^2+8x+9)}{-2(x^2+3)(x^2+3x-1)}\\
&= \frac{3}{2},
\end{align*}
so your original limit is equal to $e^{3/2}.$
