#### Answer

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.

#### Work Step by Step

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.