Question Video: Evaluating Limits of Rational Functions at Infinity | Nagwa Question Video: Evaluating Limits of Rational Functions at Infinity | Nagwa

# Question Video: Evaluating Limits of Rational Functions at Infinity Mathematics • Second Year of Secondary School

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Find lim_(π₯ β β) (7π₯Β² + 8π₯ + 4)/(5π₯Β³ + 3π₯Β²).

02:47

### Video Transcript

Find the limit as π₯ approaches β of seven π₯ squared plus eight π₯ plus four all divided by five π₯ cubed plus three π₯ squared.

The question is asking us to evaluate the limit as π₯ approaches β of a rational function. And we call this a rational function because itβs the quotient of two polynomials. Letβs call the polynomial in our numerator π of π₯ and the polynomial in our denominator π of π₯. And we know how to find the limit as π₯ approaches β of the rational function π of π₯ divided by π of π₯. First, we find the highest power of π₯ which appears in either of our polynomials, π of π₯ or π of π₯. Second, we want to divide through both our polynomials by this power of π₯. Third, we just need to evaluate the limit of this new function.

So letβs apply this method to the limit given to us in the question. First, we need to find the highest power of π₯ which appears in either π of π₯ or π of π₯. We can see the highest power of π₯ in our numerator is π₯ squared and the highest power of π₯ in our denominator is π₯ cubed. And π₯ cubed is a higher power of π₯ than π₯ squared. So weβre going to use π₯ cubed.

The next step tells us that we need to divide both π of π₯ and π of π₯ through by π₯ cubed. So we have the limit as π₯ approaches β of our rational function is equal to the limit as π₯ approaches β of seven π₯ squared plus eight π₯ plus four divided by π₯ cubed. And then we divide all of this by five π₯ cubed plus three π₯ squared divided by π₯ cubed. And, remember, weβre not changing the value of our limit because weβre dividing both the numerator and the denominator by the same value of π₯ cubed. Dividing our numerator through by π₯ cubed, we get seven over π₯ plus eight over π₯ squared plus four over π₯ cubed. And then dividing through our denominator by π₯ cubed, we get five plus three over π₯.

We now want to evaluate this new limit. And, in fact, we can do this directly. We see our limit is as π₯ is approaching β. And we can see that seven over π₯, eight over π₯ squared, four over π₯ cubed, and three over π₯ are all approaching zero as π₯ approaches β. This is because all of them have constant numerators. However, their denominators are getting larger and larger without bound. So all of the terms in our numerator approach zero. And the only term left in our denominator is the constant five. So this just tends to five. So the value of this limit is just zero divided by five, which is, of course, just equal to zero.

Therefore, by dividing our numerator and our denominator by the highest power of π₯ found in our rational function, weβve shown the limit as π₯ approaches β of seven π₯ squared plus eight π₯ plus four divided by five π₯ cubed plus three π₯ squared is equal to zero.

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