Answer
$\dfrac {10}{8}$
Work Step by Step
$\lim _{x\rightarrow \infty }\dfrac {5x^{2}+3x-1}{4x^{2}+5}=\dfrac {\dfrac {d}{dx}\left( 5x^{2}+3x-1\right) }{\dfrac {d}{dx}\left( 4x^{2}+5\right) }=\dfrac {10x+3}{8x}=\dfrac {\dfrac {d}{dx}\left( 10x+3\right) }{\dfrac {d}{dx}\left( 8x\right) }=\dfrac {10}{8}$
