The limit of $\lim\limits_{x \to \infty}\sqrt{x^2+3x-4}-x$ I tried all I know and I always get to $\infty$, Wolfram Alpha says $\frac{3}{2}$. How should I simplify it?
$$\lim\limits_{x \to \infty}\sqrt{(x^2+3x+4)}-x$$
I tried multiplying by its conjugate, taking the squared root out of the limit, dividing everything by $\sqrt{x^2}$, etc. 
Obs.: Without using l'Hôpital's.

$$ \lim_{x \rightarrow \infty} \left(x^2 + 3x + 4\right)^{ \frac{1}{2}} - x $$
$$ = \lim_{x \rightarrow \infty} \left(x^2\left(1+\frac{3}{x}+\frac{4}{x^2}\right)\right)^{ \frac{1}{2}} - x $$
$$ = \lim_{x \rightarrow \infty} x\left(1+\frac{3}{x}+\frac{4}{x^2}\right)^{ \frac{1}{2}} - x $$
Then by Taylor expansion, we get that
$$ = \lim_{x \rightarrow \infty} x\left(1+\frac{1}{2}\left(\frac{3}{x}+\frac{4}{x^2}\right)+\operatorname{o}\left(\frac{1}{x^2}\right)\right) - x $$
$$ = \lim_{x \rightarrow \infty} \frac{3}{2} + \operatorname{o}\left(\frac{1}{x}\right) = \frac{3}{2} $$
as required.
