Question Video: Finding the Limit of a Combination of Root and Polynomial Functions at a Point Using Rationalisation | Nagwa Question Video: Finding the Limit of a Combination of Root and Polynomial Functions at a Point Using Rationalisation | Nagwa

Question Video: Finding the Limit of a Combination of Root and Polynomial Functions at a Point Using Rationalisation Mathematics

Find lim_(π‘₯ β†’ ∞) (2π‘₯Β³ βˆ’ 4π‘₯Β² βˆ’ 2π‘₯ βˆ’ 7).

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Video Transcript

Find the limit as π‘₯ tends to infinity of two π‘₯ cubed minus four π‘₯ squared minus two π‘₯ minus seven.

The limit of any polynomial function as π‘₯ tends to infinity is equal to the limit of its highest-degree term, and so the limit as π‘₯ tends to infinity of two π‘₯ cubed minus four π‘₯ squared minus two π‘₯ minus seven is just the limit as π‘₯ tends to infinity of two π‘₯ cubed. This is a fact that you should be aware of, and I’ll try to explain why it’s true, at least in this case, at the end of the video.

We can now apply the constant multiple or scalar multiple property of limits, which is that the limit of a constant multiple of the function is equal to that constant multiple of the limit of the function. So the limit as π‘₯ tends to infinity of two π‘₯ cubed is two times the limit as π‘₯ tends to infinity of π‘₯ cubed.

So now we just need to find the limit as π‘₯ tends to infinity of π‘₯ to the power of three for a positive integer, and the limit as π‘₯ tends to infinity of π‘₯ to the power of 𝑛 is equal to infinity, and certainly the limit as π‘₯ tends to infinity of π‘₯ to the power of three is infinity.

We are then left with the question of what to do with two times infinity. Two times infinity is just infinity; you can’t make infinity any bigger by multiplying it by a number. The only thing that can happen is, if you multiply infinity by a negative number, then you get negative infinity instead, but as we were multiplying by two and two is definitely positive, we just get infinity.

We got the answer infinity by applying a series of rules that we just had to remember, and you may not have found that particularly satisfying, but all of these rules can be justified. For example, we claimed that the limit as π‘₯ tends to infinity of two π‘₯ cubed minus four π‘₯ squared minus two π‘₯ minus seven was just the same as the limit as π‘₯ tends to infinity of two π‘₯ cubed and, in general, that the limit of a polynomial function as π‘₯ tends to infinity only depends on its highest-order term.

While certainly using the multiplicative property of limits on the right-hand side, we can see that the product of the two limits on the right-hand side will give us the limit on the left-hand side. So here we are taking 𝑓 of π‘₯ equal to two π‘₯ cubed and 𝑔 of π‘₯ equal to two π‘₯ cubed minus four π‘₯ squared minus two π‘₯ minus seven over two π‘₯ cubed, and then 𝑓 of π‘₯, 𝑔 of π‘₯, the product of the functions becomes two π‘₯ cubed minus four π‘₯ squared minus two π‘₯ minus seven.

And we are further assuming that this still holds when 𝑐 is infinity, and we can split up the fraction in this limit, and so we get the limit as π‘₯ tends to infinity of two π‘₯ cubed over two π‘₯ cubed minus four π‘₯ squared over two π‘₯ cubed minus two π‘₯ over two π‘₯ cubed minus seven over two π‘₯ cubed.

And of course, two π‘₯ cubed over two π‘₯ cubed is just one, and we can simplify the other terms likewise, so the minus four π‘₯ squared over two π‘₯ cubed becomes minus two over π‘₯. And simplifying the next two terms, the expression of the limit becomes one minus two over π‘₯ minus one over π‘₯ squared minus seven over two π‘₯ cubed.

We can find the limit of each term individually, and if it’s not immediately obvious that all of these underlined limits are zero, you can prove this using the constant multiple or scalar multiple property and the power property to relate all these limits to the limit as π‘₯ tends to infinity of one over π‘₯, the reciprocal function.

In any case, these three limits are all zero, and so we’re just left with the limit as π‘₯ tends to infinity of one, which is of course just one, and so we see that this is the limit as π‘₯ tends to infinity of two π‘₯ cubed times one, which is just the limit as π‘₯ tends to infinity of two π‘₯ cubed.

And this is what we had to show that the limit as π‘₯ tends to infinity of two π‘₯ cubed minus four π‘₯ squared minus two π‘₯ minus seven is just equal to the limit as π‘₯ tends to infinity of two π‘₯ cubed.

The exact same argument works for a general polynomial, and so we get the polynomial property of limits that the limit as π‘₯ tends to infinity of a polynomial depends only on the highest-degree term of that polynomial, and the other properties that we used in proving that the limit as π‘₯ tends to infinity of two π‘₯ cubed minus four π‘₯ squared minus two π‘₯ minus seven equals infinity can be proved in a very similar way.

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