<s>[INST] Computing $ \lim \limits_{x \rightarrow \infty} \left(1+\frac{1}{x} \right)^{\sqrt{ax^2 +bx+c}}$ For $a \geq 0 $, the following limit 

$$ L = \lim_{x \rightarrow \infty} \left(1+\dfrac{1}{x} \right)^{\sqrt{ax^2 +bx+c}}$$

can be computed by applying L'Hopital rule as follows
$$L = \exp \left(\lim_{x \rightarrow \infty} \frac{\log \left(1+\dfrac{1}{x} \right)}{\frac{1}{\sqrt{ax^2 +bx+c}}}\right) = \exp \left(\lim_{x \rightarrow \infty} \dfrac{\frac{-1}{x(x+1)}}{\frac{-(2ax+b)}{2(ax^2 +bx+c)^{3/2}}}\right) =$$
$$ = \exp \left(\lim_{x \rightarrow \infty} \dfrac{2(ax^2 +bx+c)^{3/2}}{x(x+1)(2ax+b)}\right) =  \exp \left(\lim_{x \rightarrow \infty} \dfrac{2\left(a+ \frac{b}{x}+\frac{c}{x^2}\right)^{3/2}}{1\cdot\left(1+\frac{1}{x}\right)\left(2a+\frac{b}{x}\right)}\right) = \exp(\sqrt{a}).$$
I am wondering if exists another method to compute this limit.
Thanks for any hint!
 [/INST] I unusually use this general method:
Let $\displaystyle \lim_{x \to x_o}u(x)=1$ and $\displaystyle \lim_{x\to x_o}V(x)=\infty$, then
$\displaystyle \lim_{x \to x_o}u^V=\lim_{x \to x_o}\Big[ \Big(1+(u-1) \Big)^{\frac{1}{u-1}}\space \Big]^{(u-1)V}=\Big[ \lim_{x\to x_o}\Big( 1+(u-1)\Big)^{\frac{1}{u-1} \space} \Big]^{\displaystyle\lim_{x\to x_o}(u-1)V}=e^{\displaystyle\lim_{x\to x_o}(u-1)V}$
Now let's apply what we got above to our problem. Here $ \{ u(x) = 1+\frac{1}{x}\} \to 1 $ and $\{V(x)=\sqrt{ax^2+bx+c}\} \to \infty$ as $x\to \infty$.
$\displaystyle \lim_{x\to \infty}\left(1+\frac{1}{x} \right)^{\sqrt{ax^2+bx+c}}=[1^\infty]=\exp \left( \lim_{x\to \infty}\left( 1+\frac{1}{x}-1\right)\sqrt{ax^2+bx+c}\right)=\exp \left( \lim_{x\to \infty}\frac{1}{x}\sqrt{ax^2+bx+c}\right)=\exp \left(\lim_{x\to\infty}\sqrt{a+\frac{b}{x}+\frac{c}{x^2}}\right)=\exp \sqrt a=e^{\sqrt a}$
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