<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] 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 $e^{2}$
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