# Video: Finding the Limit of a Rational Function at Infinity

Find lim_(π₯ β β) (β2π₯Β³ + 7π₯Β² + 8π₯ + 2)/(4π₯β΄ + π₯Β³ β 2π₯Β² β 6π₯ + 7).

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### Video Transcript

Find the limit as π₯ tends to infinity of negative two π₯ cubed plus seven π₯ squared plus eight π₯ plus two all over four π₯ to the four plus π₯ cubed minus two π₯ squared minus six π₯ plus seven.

Our first step is to divide both numerator and denominator by the highest power of π₯ that we can see. In this case, this is π₯ to the four. And having done this, we can split up the fractions in the numerator and denominator. And now we can use exponent laws to simplify the terms. So, for example, negative two π₯ cubed over π₯ to the four becomes negative two over π₯. Having done this, we can use the quotient property of limits which is that the limit of a quotient is equal to the quotient of the limits as long as the denominator is nonzero.

Now we can evaluate the limits in the numerator and denominator separately. So starting with the numerator, we use the fact that the limit of a sum of functions is equal to the sum of limits of the functions. And this isnβt just true for the sum of two functions. Itβs true for the sum of any number of functions. Now we have many more limits in our numerator. But hopefully, each of them is simpler to evaluate. We also use the constant multiple property of limits, that the limit of a constant multiple of the function is equal to that constant multiple of the limit of the function. For example, we can use this property to rewrite the first limit in the numerator, taking πΎ to be negative two and π of π₯ to be one over π₯. This becomes negative two times the limit as π₯ tends to infinity of one over π₯. The negative two comes outside the limit.

We make this change and do the same for the other terms in the numerator. So now all the limits in the numerator are of the form: limit as π₯ tends to infinity of one over π₯ to the power of π, for some whole number π. And we can use the power property of limits to write all these limits in terms of limits of the reciprocal function. With π of π₯ being the reciprocal function one over π₯ and π being two, the highlighted limit becomes the limit as π₯ tends to infinity of one over π₯ squared.

And we can do something similar for the subsequent two terms. So now all four limits in the numerator are the limit as π₯ tends to infinity of one over π₯. And yet another property of limits tells us the value of this limit. The limit as π₯ tends to infinity of the reciprocal function one over π₯ is zero. This limit is zero. And so negative two times this limit is zero. And so we can get rid of this term. Itβs a similar story for the next term. The limit is zero. And so the square of the limit is zero. And so seven times the square of the limit is zero. And we get rid of this term too. Itβs exactly the same story with the other two terms. Theyβre both zero. And so our entire numerator is just zero.

We turn our attention to the denominator. And we apply the sum, constant multiple, and power properties as with the numerator to get this. As in the numerator, all the terms involving the limit as π₯ tends to infinity of the reciprocal function one over π₯ vanish. But unlike in the numerator, we still have a term left, the limit as π₯ tends to infinity of four. We need another property of limits, that the limit as π₯ tends to infinity of a constant function is constant. In fact, this is true whatever π₯ is tending to, as is the power property. Anyway, we see that the limit as π₯ tends to infinity of four is then just four. And so our fraction becomes zero over four.

Itβs good news that the denominator is nonzero. Because otherwise, weβd have an indeterminate form. And the quotient property that we used earlier is only valid if the value of the limit in the denominator is nonzero. And as zero over four is just zero, the value of the limit that we were looking for, the limit as π₯ tends to infinity of negative two π₯ cubed plus seven π₯ squared plus eight π₯ plus two all over four π₯ to the four plus π₯ cubed minus two π₯ squared minus six π₯ plus seven, is just zero.

The main trick to solving this question was dividing both numerator and denominator by π₯ to the four, which was the highest power of π₯ that we could see. After doing this, we just had to apply lots of the properties of limits: the quotient, sum, constant multiple, and power properties. And then used two known limits: the limit as π₯ tends to infinity of the reciprocal function and the limit of a constant function. We can use the same techniques to prove a more general result, that the limit as π₯ tends to infinity of a rational function, whose denominator has a greater degree than its numerator, is zero.