Answer
The value of $\underset{x\to 5}{\mathop{\lim }}\,\frac{2{{x}^{2}}-x+4}{x-1}$ is $\frac{49}{4}$.
Work Step by Step
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 $\frac{49}{4}$
