<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] Hint 
$$A=\left(1+\dfrac{1}{x} \right)^{\sqrt{ax^2 +bx+c}}\implies \log(A)={\sqrt{ax^2 +bx+c}}\log\left(1+\dfrac{1}{x} \right)$$ Now, use Taylor series $$\log\left(1+\dfrac{1}{x} \right)=\frac{1}{x}-\frac{1}{2 x^2}+O\left(\frac{1}{x^3}\right)$$ and you will arrive to the result (not only the limit but also how it is approached).
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