Before 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.

Of course, even if we had tried to plug in the point we would have run into problems as \(g\left( { - 2} \right)\) doesn’t exist and this tell us all we need to know. As noted in the notes for this section if either the function or the limit do not exist then the function is not continuous at the point. Therefore, we can see that the function is not continuous at \(z = - 2\).

For practice you might want to verify that,

\[\mathop {\lim }\limits_{z \to \, - {2^ - }} g\left( z \right) = \infty \hspace{0.5in}\hspace{0.25in}\mathop {\lim }\limits_{z \to \, - {2^ + }} g\left( z \right) = - \infty \]

and so \(\mathop {\lim }\limits_{z \to \, - 2} g\left( z \right)\) also doesn’t exist.

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).

Therefore, because we can’t just plug the point into the function, 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.

Here is the work for this part.

\[\begin{align*}\mathop {\lim }\limits_{z \to 0} g\left( z \right) = \mathop {\lim }\limits_{z \to 0} \frac{6}{{{z^2} - 3z - 10}} = \frac{{\mathop {\lim }\limits_{z \to 0} 6}}{{\mathop {\lim }\limits_{z \to 0} \left( {{z^2} - 3z - 10} \right)}} & = \frac{{\mathop {\lim }\limits_{z \to 0} 6}}{{\mathop {\lim }\limits_{z \to 0} {z^2} - 3\mathop {\lim }\limits_{z \to 0} z - \mathop {\lim }\limits_{z \to 0} 10}}\\ & = \frac{6}{{{0^2} - 3\left( 0 \right) - 10}} = g\left( 0 \right)\end{align*}\]

So, we can see that \(\mathop {\lim }\limits_{z \to 0} g\left( z \right) = g\left( 0 \right)\) and so the function is continuous at \(z = 0\).

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). Although there is also of course the problem here that \(g\left( 5 \right)\) doesn’t exist and so we couldn’t plug in the value even if we wanted to.

This also tells us what we need to know however. As noted in the notes for this section if either the function of the limit do not exist then the function is not continuous at the point. Therefore, we can see that the function is not continuous at \(z = 5\).

For practice you might want to verify that,

\[\mathop {\lim }\limits_{z \to {5^ - }} g\left( z \right) = - \infty \hspace{0.5in}\hspace{0.25in}\mathop {\lim }\limits_{z \to {5^ + }} g\left( z \right) = \infty \]

and so \(\mathop {\lim }\limits_{z \to 5} g\left( z \right)\) also doesn’t exist.