## Precalculus (6th Edition) Blitzer

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.”
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.