# Horizontal Asymptote of Strange Function

What is the horizontal asymptote as x approaches positive infinity of $\sqrt{4x^2 + 5x} - \sqrt{4x^2 + x}$? The horizontal asymptote is in the form $y = k$. Find $k$.

$$\sqrt{4x^2 + 5x} - \sqrt{4x^2 + x}=\left(\sqrt{4x^2 + 5x} - \sqrt{4x^2 + x}\right)\frac{\sqrt{4x^2 + 5x} + \sqrt{4x^2 + x}}{\sqrt{4x^2 + 5x} + \sqrt{4x^2 + x}}\\=\frac{4x^2+5x-(4x^2+x)}{\sqrt{4x^2 + 5x} + \sqrt{4x^2 + x^2}}=\frac{4x}{x\sqrt{4 + 5/x} + x\sqrt{4 + 1/x}}\\=\frac{4}{\sqrt{4 + 5/x} + \sqrt{4 + 1/x}}$$
$$\lim_{x \rightarrow +\infty}(\sqrt{4x^2 + 5x} - \sqrt{4x^2 + x})=\lim_{x \rightarrow +\infty}\frac{4}{\sqrt{4 + 5/x} + \sqrt{4 + 1/x}}=\frac{4}{\sqrt{4}+\sqrt{4}}=1$$
You could go slightly further and see not only the limit but also how the function approaches this limit. Let me rewrite $$f(x)=\sqrt{4x^2 + 5x} - \sqrt{4x^2 + x}=2x\Big(\sqrt{1+\frac{5}{4x}}-\sqrt{1+\frac{1}{4x}}\Big)$$ Now, remember that when $y$ is small $$\sqrt{1+y}=1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)$$ So $$f(x)\approx 2x\Big[\Big(1+\frac{5}{8 x}-\frac{25}{128 x^2}\Big)-\Big(1+\frac{1}{8 x}-\frac{1}{128 x^2} \Big)\Big]=2x\Big(\frac{1}{2 x}-\frac{3}{16 x^2}\Big)=1-\frac{3}{8 x}$$