Consider the function:


The numerator of the function can be factorised as:


Then we can cancel the \(x+6\) from numerator and denominator and we are left with:


However, we are only able to cancel the \(x+6\) term if \(x\ne -6\). If \(x=-6\), then the denominator becomes \(\text{0}\) and the function is not defined. This means that the domain of the function does not include \(x=-6\). But we can examine what happens to the values for \(y\) as \(x\) gets closer to \(-\text{6}\). The list of values shows that as \(x\) gets closer to \(-\text{6}\), \(y\) gets closer and closer to \(-\text{8}\).









































The graph of this function is shown below. The graph is a straight line with slope \(\text{1}\) and \(y\)-intercept \(-\text{2}\), but with a hole at \(x=-6\). As \(x\) approaches \(-\text{6}\) from the left, the \(y\)-value approaches \(-\text{8}\) and as \(x\) approaches \(-\text{6}\) from the right, the \(y\)-value approaches \(-\text{8}\). Since the function approaches the same \(y\)-value from the left and from the right, the limit exists.


[Attributions and Licenses]

This is a lesson from the tutorial, Differential Calculus and you are encouraged to log in or register, so that you can track your progress.

Log In

Share Thoughts