Precalculus (6th Edition) Blitzer

The function f\left( x \right)=\left\{ \begin{align} & 2-x\text{ if }x<1 \\ & \text{1 if }x=1\text{ } \\ & {{x}^{2}}\text{ if }x>1 \end{align} \right. is continuous at $1$.
Consider the function f\left( x \right)=\left\{ \begin{align} & 2-x\text{ if }x<1 \\ & \text{1 if }x=1\text{ } \\ & {{x}^{2}}\text{ if }x>1 \end{align} \right.. Find the value of $f\left( x \right)$ at $a=1$. From the definition of the function, $f\left( 1 \right)=1$ The function is defined at the point $a=1$. Now find the value of $\,\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)$. First find the left hand limit of $\,f\left( x \right)$. The function $f\left( x \right)=2-x$ when $x<1$. $\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=2-1=1$ Now find the right hand limit of $\,f\left( x \right)$. The function $f\left( x \right)={{x}^{2}}$ when $x>1$. $\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)={{1}^{2}}=1$ Since the left hand limit and right hand limit are equal, that is $\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)=1=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ Thus, $\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)=1$ From the above steps, $\underset{x\to 1}{\mathop{\lim }}\,f\left( x \right)=1=f\left( 1 \right)$ Thus, the function satisfies all the properties of being continuous. Hence, the function f\left( x \right)=\left\{ \begin{align} & 2-x\text{ if }x<1 \\ & \text{1 if }x=1\text{ } \\ & {{x}^{2}}\text{ if }x>1 \end{align} \right. is continuous at $1$.