## Precalculus (6th Edition) Blitzer

The statement, “If $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=L$, then the value of f at a is equal to L.” is false.
Consider the provide limit notation $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=L$. Here, $f$ is any function defined on some open interval containing the number $a$. The function f may or may not be defined at a. The notation $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=L$ means that as x gets closer to a, but remains unequal to a, the corresponding value of $f\left( x \right)$ gets closer to L but is never equal to L. Consider the limit notation, $\underset{x\to 2}{\mathop{\lim }}\,\left( {{x}^{2}}+1.9 \right)=6$ To solve the notation $\underset{x\to 2}{\mathop{\lim }}\,\left( {{x}^{2}}+1.9 \right)=6$, substitute $x=2$ in the limit notation $\underset{x\to 2}{\mathop{\lim }}\,\left( {{x}^{2}}+1.9 \right)=6$. Therefore, \begin{align} \underset{x\to 2}{\mathop{\lim }}\,\left( {{x}^{2}}+1.9 \right)\overset{?}{\mathop{=}}\,6 & \\ \left( {{2}^{2}}+1.9 \right)\overset{?}{\mathop{=}}\,6 & \\ \left( 4+1.9 \right)\overset{?}{\mathop{=}}\,6 & \\ 5.9\approx 6 & \\ \end{align} Hence, it is clear that at $x=2$ the value of $f\left( x \right)$ is not exactly 6. Therefore, the provided statement, “If $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=L$, then the value of f at a is equal to L.” is false.