$a)\underset{x\to 2}{\lim }\frac{x^3-3}{x-3}=\frac{2^3-3}{2-3}=\frac{5}{-1}=-5$  

 

$b)\underset{x\to -3}{\lim }\frac{x^3-x}{2x^2-10}=\frac{{\left(-3\right)}^3-\left(-3\right)}{2{\left(-3\right)}^2-10}=\frac{-27+3}{18-10}=\frac{-24}{8}=-3$  

 

$c)\underset{x\to -2}{\lim }\frac{4x^4-4x^2+1}{2x^2-1}=\frac{4\cdot {\left(-2\right)}^4-4\cdot {\left(-2\right)}^2+1}{2\cdot {\left(-2\right)}^2-1}=\frac{64-16+1}{7}=\frac{49}{7}=7$