Paul's Online Notes
Home / Calculus I / Limits / Computing Limits
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Section 2.5 : Computing Limits

13. Evaluate $$\displaystyle \mathop {\lim }\limits_{t \to \, - 1} \frac{{t + 1}}{{\left| {t + 1} \right|}}$$, if it exists.

Hint : Recall the mathematical definition of the absolute value function and that it is in fact a piecewise function.
Show Solution

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.