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.

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 $\lim\limits_{x \to \infty} y = e^2$.
