Answer
Yes, the function $ f\left( x \right)={{x}^{2}}-5x+6$ is continuous at $4$.
Work Step by Step
Consider the function $ f\left( x \right)={{x}^{2}}-5x+6$,
First check whether the function is defined at the point $ a $ or not.
Find the value of $ f\left( x \right)$ at $ a=4$,
$\begin{align}
& f\left( 4 \right)={{\left( 4 \right)}^{2}}-5\left( 4 \right)+6 \\
& =16-20+6 \\
& =2
\end{align}$
The function is defined at the point $ a=4$.
Now find the value of $\,\underset{x\to 4}{\mathop{\lim }}\,{{x}^{2}}-5x+6$,
$\begin{align}
& \,\underset{x\to 4}{\mathop{\lim }}\,f\left( x \right)=\,\,\underset{x\to 4}{\mathop{\lim }}\,\left( {{x}^{2}}-5x+6 \right) \\
& ={{\left( 4 \right)}^{2}}-5\left( 4 \right)+6 \\
& =16-20+6 \\
& =2
\end{align}$
Thus, $\,\underset{x\to 4}{\mathop{\lim }}\,\left( {{x}^{2}}-5x+6 \right)=2$
Now check whether $\,\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)$ or not.
From the above, $\,\underset{x\to 4}{\mathop{\lim }}\,\left( {{x}^{2}}-5x+6 \right)=2\text{ and }f\left( 4 \right)=2$
Therefore, $\,\underset{x\to 4}{\mathop{\lim }}\,f\left( x \right)=f\left( 4 \right)$
Thus, the function satisfies all the properties of being continuous.
Hence, the function $ f\left( x \right)={{x}^{2}}-5x+6$ is continuous at $4$.
