<s>[INST] 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.
 [/INST] \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*}
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