Answer
$a)0;b)0$
Work Step by Step
$a)\lim _{x\rightarrow \infty }\dfrac {4x-3}{5x^{2}+1}=\dfrac {\dfrac {4}{x}-\dfrac {3}{x^{2}}}{5+\dfrac {1}{x^{2}}}=\dfrac {0-0}{5+0}=\dfrac {0}{5}=0$
$b)\lim _{x\rightarrow \infty }\dfrac {4x-3}{5x^{2}+1}=\dfrac {\dfrac {d}{dx}\left( 4x-3\right) }{\dfrac {d}{dx}\left( 5x^{2}+1\right) }=\dfrac {4}{10x}=\dfrac {\dfrac {d}{dx}\left( 4\right) }{\dfrac {d}{dx}\left( 10x\right) }=\dfrac {0}{4}=0$
