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!

I would start with 
$$\left(1+\frac1x\right)^{\sqrt{ax^2+bx+c}} = \left(1+\frac1x\right)^{\sqrt a x \cdot\sqrt{1+\frac{b}{ax} + \frac{x}{ax^2}}} =\left(\left(1 + \frac1x\right)^x\right)^{\sqrt{a}\cdot \sqrt{1+\frac{b}{ax} + \frac{x}{ax^2}}}$$
Now, use the fact that, if $f$ and $g$ are continuous and all limits exist, $$\lim_{x\to\infty}f(x)^{g(x)} = \left(\lim_{x\to\infty}f(x)\right)^{\lim_{x\to\infty} g(x)}$$
