Answer
$\lim _{x\rightarrow \infty }\dfrac {4x}{20x+1}=\dfrac {1}{5};\lim _{x\rightarrow -\infty }\dfrac {4x}{20x+1}=\dfrac {1}{5};y=\dfrac {1}{5}$
Work Step by Step
$\dfrac {4x}{20x+1}=\dfrac {4x+\dfrac {1}{5}-\dfrac {1}{5}}{20x+1}=\dfrac {1}{5}-\dfrac {1}{5\left( 20x+1\right) }=\dfrac {1}{5}-\dfrac {1}{100x+5}$
$$\Rightarrow \lim _{x\rightarrow \infty }\dfrac {4x}{20x+1}=\lim _{x\rightarrow \infty }\left( \dfrac {1}{5}-\dfrac {1}{100x+5}\right) =\dfrac {1}{5}-0=\dfrac {1}{5}$$
$$\Rightarrow \lim _{x\rightarrow -\infty }\dfrac {4x}{20x+1}=\lim _{x\rightarrow -\infty }\left( \dfrac {1}{5}-\dfrac {1}{100x+5}\right) =\dfrac {1}{5}+0=\dfrac {1}{5}$$
So horizontal asymptote will be $y=\frac{1}{5}$
