Solution (a) To find the difference quotient of $f$, we begin with $f( x+h)$. $$\begin{eqnarray*} f(x+h)&=&2(x+h)^{2}-3(x+h)+1=2( x^{2}+2xh+h^{2}) -3x-3h+1\\[4pt] &=&2x^{2}+4xh+2h^{2}-3x-3h+1 \end{eqnarray*}$$

Now $$\begin{equation*} f( x+h) -f( x) = ( 2x^{2}+4xh+2h^{2}-3x-3h+1 ) - ( 2x^{2}-3x+1 ) =4xh+2h^{2}-3h \end{equation*}$$

Then, the difference quotient is $$\begin{equation*} \dfrac{f(x+h)-f(x)}{h}=\frac{4xh+2h^{2}-3h}{h}=\frac{h(4x+2h-3)}{h}=4x+2h-3, \hbox{ }h\neq 0 \end{equation*}$$

(b) $\lim\limits_{h\rightarrow 0}\dfrac{f(x+h)-f(x)}{h} =\lim\limits_{h\rightarrow 0}( 4x+2h-3) =4x+0-3=4x-3$