Answer
The complete statement is, “The limit notation $\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=L $ is called the right-hand limit. The notation means that as x gets closer to $a$ but remains greater than $a$, the corresponding value of $ f\left( x \right)$ gets closer to L.”
Work Step by Step
Consider the provided 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.
Hence, the notation $\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=L $ means that as $x$ gets closer to $a$ from the right, but remains unequal to $a$, the corresponding value of $ f\left( x \right)$ gets closer 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,
$\begin{align}
\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,\left( {{x}^{2}}+1.9 \right)\overset{?}{\mathop{=}}\,6 & \\
\left( {{2}^{2}}+1.9 \right)\overset{?}{\mathop{=}}\,6 & \\
\left( 4+1.9 \right)\overset{?}{\mathop{=}}\,6 & \\
5.9\approx 6 & \\
\end{align}$
Thus, it is clear that, the limit notation $\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=L $ is called the right-hand limit. The notation means that as x gets closer to $a$ but remains greater than a, the corresponding value of $ f\left( x \right)$ gets closer to L.
