Answer
$\dfrac {1}{2}=0.5$
Work Step by Step
$\lim _{x\rightarrow \infty }\dfrac {\sqrt {x^{2}-1}}{2x-1}=\dfrac {x\sqrt {1-\dfrac {1}{x^{2}}}}{2x-1}=\dfrac {\sqrt {1-\dfrac {1}{x^{2}}}}{2-\dfrac {1}{x}}=\dfrac {\sqrt {1-0}}{2-0}=\dfrac {1}{2}=0.5$
