Answer
a)$0$
b)$-\frac{2}{3}$
c)$-\infty$
Work Step by Step
a)$\lim _{x\rightarrow \infty }\dfrac {3-2x}{3x^{3}-1}=\dfrac {\dfrac {3}{x^{3}}-\dfrac {2x}{x^{3}}}{\dfrac {3x^{3}}{x^{3}}-\dfrac {1}{x^{3}}}=\dfrac {\dfrac {3}{x^{3}}-\dfrac {2}{x^{2}}}{3-\dfrac {1}{x^{3}}}=\dfrac {0-0}{3-0}=0$
b)$\lim _{x\rightarrow \infty}\dfrac {3-2x}{3x-1}=\dfrac {\dfrac {3}{x}-\dfrac {2x}{x}}{\dfrac {3x}{x}-\dfrac {1}{x}}=\dfrac {\dfrac {3}{x}-2}{3-\dfrac {1}{x}}=\dfrac {0-2}{3-0}=-\dfrac {2}{3}$
c)$\lim _{x\rightarrow 0}\dfrac {3-2x^{2}}{3x-1}=\dfrac {\dfrac {3}{x}-\dfrac {2x^{2}}{x}}{\dfrac {3x}{x}-\dfrac {1}{x}}=\dfrac {\dfrac {3}{x}-2x}{3-\dfrac {1}{x}}=\dfrac {0-\infty }{3-0}=-\dfrac {\infty }{3}=-\infty $
