Section 2.9 : Continuity
6. Using only Properties 1- 9 from the Limit Properties section, one-sided limit properties (if needed) and the definition of continuity determine if the following function is continuous or discontinuous at (a) \(t = - 2\), (b) \(t = 10\)?
\[h\left( t \right) = \left\{ {\begin{array}{rl}{{t^2}}&{t < - 2}\\{t + 6}&{t \ge - 2}\end{array}} \right.\]Show All Solutions Hide All Solutions
a Show SolutionBefore starting off with the solution to this part notice that we CAN NOT do what we’ve commonly done to evaluate limits to this point. In other words, we can’t just plug in the point to evaluate the limit. Doing this implicitly assumes that the function is continuous at the point and that is what we are being asked to determine here.
Therefore, the only way for us to compute the limit is to go back to the properties from the Limit Properties section and compute the limit as we did back in that section. We won’t be putting all the details here so if you need a little refresher on doing this you should go back to the problems from that section and work a few of them.
Also notice that for this part we have the added complication that the point we’re interested in is also the “cut-off” point of the piecewise function and so we’ll need to take a look at the two one sided limits to compute the overall limit and again because we are being asked to determine if the function is continuous at this point we’ll need to resort to basic limit properties to compute the one-sided limits and not just plug in the point (which assumes continuity again…).
Here is the work for this part.
\[\begin{array}{c}\mathop {\lim }\limits_{t \to \, - {2^ - }} h\left( t \right) = \mathop {\lim }\limits_{t \to \, - {2^ - }} {t^2} = {\left( { - 2} \right)^2} = 4\\ \mathop {\lim }\limits_{t \to \, - {2^ + }} h\left( t \right) = \mathop {\lim }\limits_{t \to \, - {2^ + }} \left( {t + 6} \right) = \mathop {\lim }\limits_{t \to \, - {2^ + }} t + \mathop {\lim }\limits_{t \to \, - {2^ + }} 6 = - 2 + 6 = 4\end{array}\]So we can see that \(\mathop {\lim }\limits_{t \to \, - {2^ - }} h\left( t \right) = \mathop {\lim }\limits_{t \to \, - {2^ + }} h\left( t \right) = 4\) and so \(\mathop {\lim }\limits_{t \to \, - 2} h\left( t \right) = 4\).
Next, a quick computation shows us that \(h\left( { - 2} \right) = - 2 + 6 = 4\) and so we can see that \[\mathop {\lim }\limits_{t \to \, - 2} h\left( t \right) = h\left( { - 2} \right)\] and so the function is continuous at \(t = - 2\).
b Show Solution
For justification on why we can’t just plug in the number here check out the comment at the beginning of the solution to (a).
For this part we can notice that because there are values of \(t\) on both sides of \(t = 10\) in the range \(t \ge - 2\) we won’t need to worry about one-sided limits here. Here is the work for this part.
Here is the work for this part.
\[\mathop {\lim }\limits_{t \to \,10} h\left( t \right) = \mathop {\lim }\limits_{t \to \,10} \left( {t + 6} \right) = \mathop {\lim }\limits_{t \to \,10} t + \mathop {\lim }\limits_{t \to \,10} 6 = 10 + 6 = h\left( {10} \right)\]So, we can see that \(\mathop {\lim }\limits_{t \to \,10} h\left( t \right) = h\left( {10} \right)\) and so the function is continuous at \(t = 10\).