### Video Transcript

Consider the function π of π₯ is equal to π₯ squared plus two π₯ minus eight divided by π₯ minus two when π₯ is not equal to two and two π when π₯ is equal to two. Find π such that π of π₯ is continuous for all values of π₯.

The first thing we can do for this question is to recall the conditions for a continuous function. We say that a function is continuous at a point π₯ equals π if the limit as π₯ approaches π of π of π₯ is equal to the function evaluated at π₯ equals π, π of π. This question is asking us to find the value of π such that our function is continuous for all points or all values of π₯. Our function π of π₯ is almost entirely defined by the first subfunction, π₯ squared plus two π₯ minus eight all divided by π₯ minus two. Weβll choose to call this subfunction β of π₯. β of π₯ takes the form of a rational function. It should be clear to us that this quotient will have a nonzero denominator at all values, where π₯ is not equal to two. Luckily, π₯ equals two is the only point at which π of π₯ is not defined by this subfunction.

Given this information, we can conclude that β of π₯ will behave like π normal polynomial at all points where π₯ is not equal to two. A fact that we can then use to help us is that polynomial functions are continuous on the entire set of real numbers. Since π of π₯ is defined by β of π₯ when π₯ is not equal to two and β of π₯ is continuous when π₯ is not equal to two, it therefore follows that π of π₯ is continuous when π₯ is not equal to two. Great! Weβre now happy with the fact that of π₯ is continuous for all real numbers aside from π₯ equals two.

Letβs now deal with this. Letβs now go back to our continuity condition. We need to make sure this is satisfied for π₯ equals two in order for π of π₯ to be continuous for all values of π₯. Now remember the limit as π₯ approaches two of our function π of π₯ concerns values of π₯ which are arbitrarily close to two, but not equal to two. This means that weβll be working with our first subfunction β of π₯ when looking for the limit as π₯ approaches two of π of π₯.

The right-hand side of our equation for the continuity condition is fairly trivial since π of two is simply two π. And weβll be setting this value of π as desired. The left-hand side of the equationβthe limitβis where all the legwork happens. So letβs work on this. As we just said, the limit as π₯ approaches two of π of π₯ is given by the limit as π₯ approaches two of our first subfunction, β of π₯. Letβs use the expression that we have.

To find this limit, the first thing we may think to do is take a direct substitution approach. If we were to do this, substituting two for π₯, and then we were to simplify our answer, we would be left with the indeterminate form of zero over zero. This is not very helpful to us. And in fact, we already knew that π₯ equals two would give a denominator of zero for this subfunction. Instead of this, weβre gonna take a different approach, evaluating our limit by factorisation. Letβs rewind back to β of π₯.

With a little bit of inspection, we might notice that the numerator of β of π₯ can be factorised to π₯ minus two times π₯ plus four. If we were to then cancel the common factor of π₯ minus two from the top and bottom half of our quotient, we get π₯ plus four. But here, we should be a little bit careful not to call this π₯ plus four β of π₯. Letβs call it something different. How about π of π₯? Now, you might be wondering why weβve defined these two things in different ways since clearly they are equal. And although this is nearly true, β of π₯ is equal to π of π₯, but only if π₯ is not equal to two. Remember we proved that π₯ equals two is not in the domain of β of π₯ since this gives us a divide by zero. Clearly, π₯ equals two is in the domain of π of π₯.

Now, letβs consult another general rule to move forward. This rule says that the limit as π₯ approaches π of some function, which weβve called β of π₯, is equal to the limit as π₯ approaches π of some other function π of π₯ if these two functions are equal at all values where π₯ is not equal to π. Again, remember this is because limits concern values of π₯ which are arbitrarily close to π₯ equals π, but not π₯ equals π itself. Here, we see the condition for this general rule is the exact same condition to the one weβve just observed.

This now allows us to move forward with our question in the following way. Here, we see that the limit as π₯ approaches two of β of π₯ is equal to the Limit as π₯ approaches two of π of π₯. And we remember the steps we took to get here. Since π₯ equals two is in the domain of π of π₯, we can now use the direct substitution approach. Subbing in, we get the limit is equal to two plus four which is of course six. Now, finally, once again, we get back to our condition for continuity.

Now that we know the left-hand side of this equation is equal to six, we must also set the right-hand side of the equation to be six. This means that in order for our continuity condition to be satisfied, π of two must be equal to six. Since π of π₯ is defined by the second subfunction two π when π₯ equals two, we know that π of two is two π, as we stated earlier. Simply by dividing both sides of this equation by two, we reach the result that π must equal three in order for the continuity condition to be satisfied when π₯ equals two.

In fact, if π equals three, given our previous reasoning, we know that the continuity condition is satisfied over the entire set of real numbers, so for all values of π₯. This means that we have answered our question. And we found the value for π such that our function π of π₯ is continuous for all values of π₯.