K= $\lim_{x \to 0} $ $\frac{\sqrt{4x+1}-1}{x^2-3x}$
=$\lim_{x \to 0} $ $\frac{(\sqrt{4x+1}-1)(\sqrt{4x+1}+1)}{x(x-3)(\sqrt{4x+1}+1)}$
=$\lim_{x \to 0} $ $\frac{4x}{x(x-3)(\sqrt{4x+1}+1)}$
=$\lim_{x \to 0} $ $\frac{4}{(x-3)(\sqrt{4x+1}+1)}$
=$\frac{4}{-6}$
=$-\frac{2}{3}$