Answer
$a)\dfrac {5}{3};b)\dfrac {5}{3}$
Work Step by Step
$a)\lim _{x\rightarrow \infty }\dfrac {5x^{2}-3x+1}{3x^{2}-5}=\dfrac {5-\dfrac {3}{x}+\dfrac {1}{x^{2}}}{3-\dfrac {5}{x^{2}}}=\dfrac {5-0+0}{3-0}=\dfrac {5}{3}$
$b)\lim _{x\rightarrow \infty }\dfrac {5x^{2}-3x+1}{3x^{2}-5}=\dfrac {\dfrac {d}{dx}\left( 5x^{2}-3x+1\right) }{\dfrac {d}{dx}\left( 3x^{2}-5\right) }=\dfrac {10x-3}{6x}=\dfrac {\dfrac {d}{dx}\left( 10x-3\right) }{\dfrac {d}{dx}\left( 6x\right) }=\dfrac {10}{6}=\dfrac {5}{3}$
