Question Video: Finding the Limit of a Polynomial Function Mathematics • Higher Education

Determine lim_(π‘₯ β†’ βˆ’5) (βˆ’9π‘₯Β² βˆ’ 6π‘₯ βˆ’ 9).

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Video Transcript

Determine the limit as π‘₯ tends to negative five of negative nine π‘₯ squared minus six π‘₯ minus nine.

The first step is to write down this limit again, and we can apply the difference property of limits, which says that the limit of a difference of functions is equal to the difference of the limits of those functions. So setting 𝑓 of π‘₯ equal to negative nine π‘₯ squared minus six π‘₯ and 𝑔 of π‘₯ equal to nine, we get that the limit as π‘₯ tends to negative five of negative nine π‘₯ squared minus six π‘₯ minus nine is equal to the limit as π‘₯ tends to negative five of negative nine π‘₯ squared minus six π‘₯ minus the limit as π‘₯ tends to negative five of nine.

So now in the place of one very complicated limit, we have to simpler limits. And in fact we can split up the first limit as well into the difference of two, limits setting 𝑓 of π‘₯ equals negative nine π‘₯ squared this time and 𝑔 of π‘₯ equal to six π‘₯. And so applying this. we get the limit as π‘₯ tends to negative five of negative nine π‘₯ squared minus the limit as π‘₯ tends to negative five of six π‘₯ minus the limit as π‘₯ tends to negative five of nine. And comparing this with what we started with, we see that we could’ve done this in one step.

The limit of a sum or difference of any number of terms is equal to the sum or difference as appropriate of the limits of those terms. We can now move on to finding the limit of each term individually. And so for the first term, the limit as π‘₯ tends to negative five of negative nine π‘₯ squared, we can apply the scalar multiple or constant multiple property, which says that the limit of a constant multiple of the function is equal to that constant multiple times the limit of that function.

So taking π‘˜ equal to negative nine and 𝑓 of π‘₯ equal to π‘₯ squared, we get that the limit as π‘₯ tends to negative five of negative nine π‘₯ squared is equal to negative nine times the limit as π‘₯ tends to negative five of π‘₯ squared. We can apply the same property to the limit as π‘₯ tends to negative five of six π‘₯. This becomes six times the limit as π‘₯ tends to negative five of π‘₯. And finally, we bring down the limit as π‘₯ tends to negative five of nine.

We can use the power property to further rewrite the first term. So the limit of a power of a function is equal to that power of the limit of the function. So setting 𝑓 of π‘₯ equal to π‘₯ and 𝑛 equal to two, the first term becomes negative nine times the limit as π‘₯ tends to negative five of π‘₯ squared. We carry down the other two terms unchanged from the previous line. And now we can see that the only limits we have to find are the limit of the identity function π‘₯ as π‘₯ tends to negative five.

We’ve got two of those, and the limit as π‘₯ tends to negative five of the constant function nine. And hopefully, we know what these two limits should be. The limits of the identity function π‘₯ as π‘₯ tends to 𝑐 is just 𝑐, and the limit of a constant function π‘˜ as π‘₯ tends to 𝑐 is just that constant value π‘˜. In our case, 𝑐 is equal to negative five, so the limit as π‘₯ tends to negative five of π‘₯ is just negative five. So the first two terms become negative nine times negative five squared minus six times negative five.

And taking π‘˜ to be nine, we see the limit as π‘₯ tends to negative five of nine is just nine. So finally we’ve got rid of all the limits and we have a numerical expression that we can just evaluate. Before we do that, let’s just note that this line is exactly what you would get if you took the function that we were finding the limit of and just substituted in the limit point negative five. The limit as π‘₯ tends negative five of this function is just the function evaluated as negative five.

The set of functions for which is this true is a very nice set of functions, which you will come across soon if you haven’t already, and this set of functions includes all polynomial functions. So to find the limit of a polynomial function, all you have to do is substitute in the limit point. This isn’t true for all functions, but it is certainly true for polynomial functions. Anyway let’s get on with finding our final answer. Putting this into our calculator, we get negative 204. The limit as π‘₯ tends to negative five of negative nine π‘₯ squared minus six π‘₯ minus nine is negative 204.

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