## Precalculus (6th Edition) Blitzer

Yes, the function $f\left( x \right)=2x+5$ is continuous at $1$.
Consider the function $f\left( x \right)=2x+5$, First check whether the function is defined at the point $a$ or not. Find the value of $f\left( x \right)$ at $a=1$, \begin{align} & f\left( 1 \right)=2\left( 1 \right)+5 \\ & =2+5 \\ & =7 \end{align} The function is defined at the point $a=1$. Now check whether the value of $\,\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)$ exists or not. The value of $\,\underset{x\to 1}{\mathop{\lim }}\,\left( 2x+5 \right)$, can be calculated as: \begin{align} & \,\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)=\,\underset{x\to 1}{\mathop{\lim }}\,\left( 2x+5 \right) \\ & =2\left( 1 \right)+5 \\ & =2+5 \\ & =7 \end{align} Thus, $\,\underset{x\to 1}{\mathop{\lim }}\,\left( 2x+5 \right)=7$ Now check whether $\,\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)$ or not. From the above $\,\underset{x\to 1}{\mathop{\lim }}\,\left( 2x+5 \right)=7\text{ and }f\left( 1 \right)=7$ Therefore, $\,\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)=f\left( 1 \right)$ Thus, the function satisfies all the properties of being continuous. Hence, the function $f\left( x \right)=2x+5$ is continuous at $1$.