Recall the definition of the absolute value function.

\[\left| p \right| = \left\{ {\begin{array}{*{20}{c}}p&{p \ge 0}\\{ - p}&{p < 0}\end{array}} \right.\]

So, because the function inside the absolute value is zero at \(x = 5\) we can see that,

\[\left| {x - 5} \right| = \left\{ {\begin{array}{*{20}{c}}{x - 5}&{x \ge 5}\\{ - \left( {x - 5} \right)}&{x < 5}\end{array}} \right.\]

This means that we are being asked to compute the limit at the “cut–off” point in a piecewise function and so, as we saw in this section, we’ll need to look at two one-sided limits in order to determine if this limit exists (and its value if it does exist).

\[\mathop {\lim }\limits_{x \to {5^{\, - }}} \left( {10 + \left| {x - 5} \right|} \right) = \mathop {\lim }\limits_{x \to {5^{\, - }}} \left( {10 - \left( {x - 5} \right)} \right) = \mathop {\lim }\limits_{x \to {5^{\, - }}} \left( {15 - x} \right) = 10\hspace{0.25in}{\mbox{recall }}x \to {5^ - }{\mbox{ implies }}x < 5\] \[\mathop {\lim }\limits_{x \to {5^{\, + }}} \left( {10 + \left| {x - 5} \right|} \right) = \mathop {\lim }\limits_{x \to {5^{\, + }}} \left( {10 + \left( {x - 5} \right)} \right) = \mathop {\lim }\limits_{x \to {5^{\, + }}} \left( {5 + x} \right) = 10\hspace{0.25in}{\mbox{recall }}x \to {5^ + }{\mbox{ implies }}x > 5\]

So, for this problem, we can see that,

\[\mathop {\lim }\limits_{x \to {5^{\, - }}} \left( {10 + \left| {x - 5} \right|} \right) = \mathop {\lim }\limits_{x \to 5 + } \left( {10 + \left| {x - 5} \right|} \right) = 10\]

and so the overall limit must exist and,

\[\mathop {\lim }\limits_{x \to {5}} \left( {10 + \left| {x - 5} \right|} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{10}}\]