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.

Note that
\begin{align}
\sqrt{x^2+3x-4} - x & = \left(\sqrt{x^2+3x-4} - x \right) \times \dfrac{\sqrt{x^2+3x-4} + x}{\sqrt{x^2+3x-4} + x}\\
& = \dfrac{(\sqrt{x^2+3x-4} - x)(\sqrt{x^2+3x-4} + x)}{\sqrt{x^2+3x-4} + x}\\
& = \dfrac{x^2+3x-4-x^2}{\sqrt{x^2+3x-4} + x} = \dfrac{3x-4}{\sqrt{x^2+3x-4} + x}\\
& = \dfrac{3-4/x}{\sqrt{1+3/x-4/x^2} + 1}
\end{align}
Now we get
\begin{align}
\lim_{x \to \infty}\sqrt{x^2+3x-4} - x & = \lim_{x \to \infty} \dfrac{3-4/x}{\sqrt{1+3/x-4/x^2} + 1}\\
& = \dfrac{3-\lim_{x \to \infty} 4/x}{1 + \lim_{x \to \infty} \sqrt{1+3/x-4/x^2} } = \dfrac{3}{1+1}\\
& = \dfrac32
\end{align}
