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 \(t = - 1\) we can see that,

\[\left| {t + 1} \right| = \left\{ {\begin{array}{*{20}{c}}{t + 1}&{t \ge - 1}\\{ - \left( {t + 1} \right)}&{t < - 1}\end{array}} \right.\]

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).

\[\mathop {\lim }\limits_{t \to \, - {1^{\, - }}} \frac{{t + 1}}{{\left| {t + 1} \right|}} = \mathop {\lim }\limits_{t \to \, - {1^{\, - }}} \frac{{t + 1}}{{ - \left( {t + 1} \right)}} = \mathop {\lim }\limits_{t \to \, - {1^{\, - }}} - 1 = - 1\hspace{0.25in}{\mbox{recall }}t \to - {1^ - }{\mbox{ implies }}t < - 1\] \[\mathop {\lim }\limits_{t \to \, - {1^{\, + }}} \frac{{t + 1}}{{\left| {t + 1} \right|}} = \mathop {\lim }\limits_{t \to \, - {1^{\, + }}} \frac{{t + 1}}{{t + 1}} = \mathop {\lim }\limits_{t \to \, - {1^{\, + }}} 1 = 1\hspace{0.25in}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\mbox{recall }}t \to - {1^ + }{\mbox{ implies }}t > - 1\]

So, for this problem, we can see that,

\[\mathop {\lim }\limits_{t \to \, - {1^{\, - }}} \frac{{t + 1}}{{\left| {t + 1} \right|}} = - 1 \ne 1 = \mathop {\lim }\limits_{t \to \, - {1^{\, + }}} \frac{{t + 1}}{{\left| {t + 1} \right|}}\]

and so the overall limit does not exist.