Question Video: Finding the Limit of Rational Functions at Infinity | Nagwa Question Video: Finding the Limit of Rational Functions at Infinity | Nagwa

Question Video: Finding the Limit of Rational Functions at Infinity Mathematics • Second Year of Secondary School

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Find lim_(π‘₯ β†’ ∞) 6π‘₯Β²/(π‘₯ βˆ’ 6).

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Video Transcript

Find the limit of six π‘₯ squared over π‘₯ minus six as π‘₯ approaches infinity.

This is the limit of a rational function as π‘₯ approaches infinity. And as a rational function, it is the quotient of two polynomials. And as we know something about the limits of polynomials, we might rush to use the fact that the limit of a quotient of functions is the quotient of their limits, should those limits exist. If we try applying this law, we get the limit of six π‘₯ squared as π‘₯ approaches infinity divided by the limit of π‘₯ minus six as π‘₯ approaches infinity.

Now, what is the limit of six π‘₯ squared as π‘₯ approaches infinity? Well, as π‘₯ is getting bigger and bigger, π‘₯ squared is getting bigger and bigger. And six π‘₯ squared is getting bigger and bigger or without bound. And so, this limit is infinity.

Now, what about the limit in the denominator, the limit of π‘₯ minus six as π‘₯ approaches infinity? Well, as π‘₯ approaches infinity, π‘₯ minus six is always only six steps behind. So π‘₯ minus six also goes to infinity. And so, we have infinity in the denominator also. We’re left with the indeterminate form infinity over infinity. This indeterminate form doesn’t give us any information as to what the limit we’re looking for is. The limits could be infinity. It could also be zero or, indeed, any other real number.

Just as a quick aside, you might be tempted to cancel the infinities in the numerator and denominator to get an answer of just one. But unfortunately, you can’t just divide one infinity by another and expect to get a right answer. If we get an indeterminate form like infinity over infinity or infinity minus infinity or zero over zero, this tells us that we have to find the limit in some other way. Applying the quotient law of limits straight away didn’t work. We’re going to have to be slightly more clever here.

To find the limit of a rational function at infinity or minus infinity, the trick is to divide the numerator and denominator by the highest power of the variable in the denominator first, before applying the quotient law of limits. For our rational function, the denominator is the polynomial π‘₯ minus six. And the highest power of π‘₯ in this denominator is just π‘₯ or π‘₯ to the one. Having identified the highest power of π‘₯ in the denominator, we now need to divide both numerator and denominator by this. So the numerator is now six π‘₯ squared over π‘₯. And the denominator is now π‘₯ minus six over π‘₯.

We can simplify both numerator and denominator. Six π‘₯ squared of π‘₯ is just six π‘₯. And we can split the fraction in the denominator, writing π‘₯ over π‘₯ minus six over π‘₯. And of course, π‘₯ over π‘₯ simplifies to just one. Okay, well, how does this help? Well, I claim that now when we apply the quotient law of limits, we won’t get an indeterminate form. Let’s see if I’m right.

By applying this law, we have the limit of six π‘₯ as π‘₯ approaches infinity divided by the limit of one minus six over π‘₯ as π‘₯ approaches infinity. And as before, we consider the limit in the numerator first. As π‘₯ approaches infinity, six times π‘₯ is certainly going to approach infinity. So we have infinity in the numerator.

How about in the denominator, where we have the limit of one minus six over π‘₯ as π‘₯ approaches infinity? Well, six over π‘₯ is going to get smaller and smaller as π‘₯ gets bigger and bigger. And so, one minus six over π‘₯ is going to get closer and closer to one. Of course, we could show that the limit in the denominator is one more formally by using the fact that the limit of a difference of functions is the difference of their limits, that the limit of a constant function is just that constant, that the limit of a number times a function is that number times the limit of the function, and that the limit of the reciprocal function one over π‘₯ as π‘₯ approaches infinity is zero. The value of the limit in the denominator is then one minus zero, which is one as we claimed.

Okay, so using this procedure, we found that the limit of six π‘₯ squared over π‘₯ minus six as π‘₯ approaches infinity is infinity over one. And it may not be obvious that this answer is any better than the answer of infinity over infinity that we got before. But, in this case, the thing that we’re tempted to do actually works. We can say that infinity over one is just infinity. And this is our answer. The limit of six π‘₯ squared over π‘₯ minus six as π‘₯ approaches infinity is infinity. This is because unlike infinity over infinity, infinity over one is not an indeterminate form.

The indeterminate forms that you’re likely to come across are as follows: zero over zero, infinity over infinity, zero times infinity, one to the power of infinity, infinity minus infinity, zero to the power of zero, and infinity to the power of zero. These indeterminate forms can occur when you’re trying to find the value of some limit. For example, we saw earlier that we got the indeterminate form infinity over infinity. And they’re called indeterminate forms as they give absolutely no clue as to what the limit actually is.

Naively, you might think that infinity over infinity should simplify to one. But, when we tried to find the value of our limit, we got the indeterminate form infinity over infinity. And it turns out that our limit had an infinite value. Similarly, you might think that if you get the expression zero over zero when trying to find the limit of some function, then the limit of that function is one. But again, this is not necessarily true. And the same is true for the other indeterminate forms. The obvious value is not necessarily the value of the limit which produced that indeterminate form.

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