Video: Finding the Limit of Rational Functions at Infinity

Find lim_(π‘₯ β†’ ∞) (π‘₯Β² + 3)/(8π‘₯Β³ + 9π‘₯ + 1).

02:42

Video Transcript

Find the limit of π‘₯ squared plus three all over eight π‘₯ cubed plus nine π‘₯ plus one as π‘₯ approaches infinity.

Our first thought might be to use the fact that the limit of a quotient is the quotient of the limits. This gives us the limit of π‘₯ squared plus three as π‘₯ approaches infinity over the limit of eight π‘₯ cubed plus nine π‘₯ plus one as π‘₯ approaches infinity. But we run into problems because neither limit is defined. In the numerator, as π‘₯ approaches infinity, π‘₯ squared plus three doesn’t approach any real value, it just gets bigger and bigger without bound. And the same thing happens in the denominator. As π‘₯ increases without bound, the cubic eight π‘₯ cubed plus nine π‘₯ plus one also increases without bound.

Or maybe you think both limits should be infinity. And so, the limit on left-hand side is infinity over infinity. But this, just like a zero over zero, is an indeterminate form. And it doesn’t tell us the value of our limit. We need to use a different approach.

The trick to this question is to find the highest power of π‘₯ that appears in the numerator or denominator. That’s π‘₯ cubed here. And having found this highest power, we divide both numerator and denominator by it. What do we get? π‘₯ squared divided by π‘₯ cubed is π‘₯ to the negative one. And three divided by π‘₯ cubed is three π‘₯ to the negative three. And in the denominator eight π‘₯ cubed divided by π‘₯ cubed is just eight. Nine π‘₯ divided by π‘₯ cubed is nine π‘₯ to the negative two. And one divided by π‘₯ cubed is π‘₯ to the negative three.

So, now, we just have negative powers of π‘₯ and a constant in the numerator and denominator. And as a result, when we apply this limit law, we find that the limits in the numerator and denominator do now exist. Let’s find their values. We can use the fact that the limit of a sum of functions is the sum of their limits. This allows us to find the limit of each term separately. We can also take the coefficient outside the limits.

And now, apart from one limit, which is the limit of a constant function, and whose value must therefore be eight, all the other limits have the form the limit of π‘₯ to the power of negative 𝑛 as π‘₯ approaches infinity. Where 𝑛 is, of course, greater than zero. And we know the value of such limits. The value is always zero.

So, this is zero, and this is zero, and this is zero, and this is zero. Simplifying then, our answer is zero over eight, which is, of course, just zero.

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