## Precalculus (6th Edition) Blitzer

Yes, the function $f\left( x \right)={{x}^{2}}-5x+6$ is continuous at $4$.
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 $4$.