Answer
The value of $\underset{x\to 5}{\mathop{\lim }}\,\frac{2{{x}^{2}}-7x-15}{x-5}$ is $13$.
Work Step by Step
Consider the function $ f\left( x \right)=\frac{2{{x}^{2}}-7x-15}{x-5}$,
Simplify the above function,
$\begin{align}
& f\left( x \right)=\frac{2{{x}^{2}}-7x-15}{x-5} \\
& =\frac{2{{x}^{2}}-10x+3x-15}{x-5} \\
& =\frac{2x\left( x-5 \right)+3\left( x-5 \right)}{x-5} \\
& =\frac{\left( x-5 \right)\left( 2x+3 \right)}{\left( x-5 \right)}
\end{align}$
Further simplify the above expression,
$\begin{align}
& f\left( x \right)=\frac{\left( x-5 \right)\left( 2x+3 \right)}{\left( x-5 \right)} \\
& =2x+3
\end{align}$
The function $ f\left( x \right)=2x+3$ is a polynomial.
Now, find the value of $\underset{x\to 5}{\mathop{\lim }}\,\frac{2{{x}^{2}}-7x-15}{x-5}$
$\begin{align}
& \underset{x\to 5}{\mathop{\lim }}\,\frac{2{{x}^{2}}-7x-15}{x-5}=\underset{x\to 5}{\mathop{\lim }}\,2x+3 \\
& =2\underset{x\to 5}{\mathop{\lim }}\,x+3 \\
& =2\left( 5 \right)+3 \\
& =13
\end{align}$
Thus, the value of $\underset{x\to 5}{\mathop{\lim }}\,\frac{2{{x}^{2}}-7x-15}{x-5}$ is $13$.
