Limit of $\sqrt{4x^2 + 3x} - 2x$ as $x \to \infty$ $$\lim_{x\to\infty} \sqrt{4x^2 + 3x} - 2x$$ 
I thought I could multiply both numerator and denominator by $\frac{1}{x}$, giving
$$\lim_{x\to\infty}\frac{\sqrt{4 + \frac{3}{x}} -2}{\frac{1}{x}}$$
then as x approaches infinity, $\frac{3}{x}$ essentially becomes zero, so we're left with 2-2 in the numerator and $\frac{1}{x}$ in the denominator, which I thought would mean that the limit is zero.
That's apparently wrong and I understand (algebraically) how to solve the problem using the conjugate, but I don't understand what's wrong about the method I tried to use.

I always change the variable with $y = \frac{1}{x}$ and take the limit to $y\rightarrow 0$
$$ {\rm limit} = \sqrt{\frac{4}{y^2} + \frac{3}{y}} - \frac{2}{y} = \left. \frac{\sqrt{3 y+4}-2}{y} \right|_{y\rightarrow 0} $$
No with LH rule 
$$ {\rm limit} =\left. \frac{3}{2 \sqrt{3 y+4}} \right|_{y\rightarrow 0}  = \frac{3}{4} $$
