## Precalculus (6th Edition) Blitzer

The value of $\underset{x\to 5}{\mathop{\lim }}\,\frac{2{{x}^{2}}-x+4}{x-1}$ is $\frac{49}{4}$.
Consider the function $f\left( x \right)=\frac{2{{x}^{2}}-x+4}{x-1}$, The functions $g\left( x \right)=2{{x}^{2}}-x+4$ and $h\left( x \right)=x-1$ are polynomial. To find the value of $\underset{x\to 5}{\mathop{\lim }}\,\frac{2{{x}^{2}}-x+4}{x-1}$, \begin{align} & \underset{x\to 5}{\mathop{\lim }}\,\frac{2{{x}^{2}}-x+4}{x-1}=\frac{2\underset{x\to 5}{\mathop{\lim }}\,{{x}^{2}}-\underset{x\to 5}{\mathop{\lim }}\,x+4}{\underset{x\to 5}{\mathop{\lim }}\,x-1} \\ & =\frac{2{{\left( 5 \right)}^{2}}-\left( 5 \right)+4}{\left( 5 \right)-1} \\ & =\frac{2\left( 25 \right)-5+4}{4} \\ & =\frac{50-1}{4} \end{align} $=\frac{49}{4}$ Thus, the value of $\underset{x\to 5}{\mathop{\lim }}\,\frac{2{{x}^{2}}-x+4}{x-1}$ is $\frac{49}{4}$