Find $\lim\limits_{x \to \infty} \left( \frac{\sqrt{x^2+2x-3}}{x+2} \right)^{3-2x}$ How can I find this limit?
$$\lim\limits_{x \to \infty} \bigg ( \dfrac{\sqrt{x^2+2x-3}}{x+2} \bigg )^{3-2x}$$
Firstly I thought I can use the limit:
$$\lim\limits_{x \to \infty} \bigg ( 1 + \dfrac{1}{x} \bigg )^x=e$$
by adding $1$ and subtracting $1$ from the original limit. However, since $3-2x$ $\rightarrow - \infty$ and not $+\infty$, I got nowhere. Then I tried finding the logarithm of this limit. It resulted in a $\dfrac{0}{0}$ indeterminate form, I tried L'Hospital, but again, it led me nowhere. Either I made some mistakes in the calculations, or I should use a different approach.
 A: Yes, add and subtract $1$. You will get 
$$\left[
\left(
1+\frac{\sqrt{x^2+2x-3}-x-2}{x+2}
\right)^{\frac{x+2}{\sqrt{x^2+2x-3}-x-2}}
\right]^{\frac{(\sqrt{x^2+2x-3}-x-2)(3-2x)}{x+2}}$$
The part inside the $\left[...\right]$ tends to $e$.
Then compute the limit of the exponent 
$$\begin{align}\frac{(\sqrt{x^2+2x-3}-x-2)(3-2x)}{x+2}&=\frac{((x^2+2x-3)-(x+2)^2)(3-2x)}{(\sqrt{x^2+2x-3}+x+2)(x+2)}\\
&=\frac{(-2x-7)(3-2x)}{(\sqrt{x^2+2x-3}+x+2)(x+2)}\\
&=\frac{(-2-7/x)(3/x-2)}{(\sqrt{1+2/x-3/x^2}+1+2/x)(1+2/x)}\\
&\to2\end{align}$$
Therefore, the original limit is $e^{2}$
A: \begin{align*}
&\left(\dfrac{\sqrt{x^{2}+2x-3}}{x+2}\right)^{3-2x}\\
&=\left(\dfrac{\sqrt{x^{2}+2x-3}}{\sqrt{x^{2}+2x}}\right)^{3-2x}\left(\dfrac{\sqrt{x^{2}+2x}}{x+2}\right)^{3-2x}\\
&=\left(1-\dfrac{3}{x^{2}+2x}\right)^{3/2-x}\left(\dfrac{x}{x+2}\right)^{3/2-x}\\
&=\left(1-\dfrac{3}{x^{2}+2x}\right)^{-(x^{2}+2x)(x-3/2)/(x^{2}+2x)}\left(1-\dfrac{2}{x+2}\right)^{-(x+2)(x-3/2)/(x+2)}\\
&\rightarrow 1\cdot e^{2}\\
&=e^{2}. 
\end{align*}
A: Take logarithm and substitute by $x=1/y$ to get
$$
\frac{{\left(3 \, y - 2\right)} \log\left(3 \, y + 1\right) - 2 \, {\left(3 \, y - 2\right)} \log\left(2 \, y + 1\right) + 3 \, y \log\left(-y + 1\right) - 2 \, \log\left(-y + 1\right)}{2 \, y}
=
2+O(y)
$$
as $y \to 0$, where we use $\log(1+y)=y+O(y^2)$.
A: Let $y =\bigg ( \dfrac{\sqrt{x^2+2x-3}}{x+2} \bigg )^{3-2x}$. Then,
$$\ln y = (3-2x)\ln \sqrt{\dfrac{x^2+2x-3}{(x+2)^2}} $$
$$= \frac{3-2x}2 \ln \left(1-\frac2{x+2}+O(\frac1{(x+2)^2})\right)
= \frac{\ln\left( 1-\frac2{x+2}\right)+O(\frac1{(x+2)^2})}{\frac1{-(x+2)}+O(\frac1{(x+2)^2})}$$
Therefore, 
$$\lim\limits_{x \to \infty} \ln y 
= \lim\limits_{x \to \infty}\frac{\ln\left( 1-\frac2{x+2}\right)}{\frac1{-(x+2)}} =2$$
which leads to $\lim\limits_{x \to \infty} y = e^2$.
