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

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

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

03:18

Video Transcript

Find the limit as π₯ approaches β of negative three π₯ squared over negative four π₯ squared plus eight.

In this question, we are given the rational function negative three π₯ squared over negative four π₯ squared plus eight and are required to find its limit as π₯ approaches β. Now, β is not a number, so it is not possible to evaluate this limit using direct substitution. Therefore, we will have to think of other methods to find this limit. Firstly, recall that we can pull constants out of a limit. Using this, we can factor the constant negative one out of our limit. This step makes our limit somewhat simpler to deal with.

Next, we divide each term in the numerator and denominator of our limit by the highest power of π₯ in the expression, which is π₯ squared. Doing so, we obtain negative one timesed by the limit as π₯ approaches β of three over negative four plus eight over π₯ squared. Now, using the fact that the limit of a quotient of functions is the quotient of their limits if the limit of the denominator is nonzero. We obtain negative one timesed by the limit as π₯ approaches β of three over the limit as π₯ approaches β of negative four plus eight over π₯ squared.

We must ensure that the limit in the denominator is nonzero. And this is what we will do next. Using the fact that the limit of a sum of functions is the sum of their limits, we can write the denominator as the limit as π₯ approaches β of negative four plus the limit as π₯ approaches β of eight over π₯ squared. Using the fact that the limit of a constant is just that constant, we have that the limit as π₯ approaches β of three is three. And the limit as π₯ approaches β of negative four is negative four.

Using the fact that we can pull constants out of a limit, we have that the limit as π₯ approaches β of eight over π₯ squared is equal to eight times the limit as π₯ approaches β of one over π₯ squared. Now, note that one over π₯ squared is equal to one over π₯ all squared. Now, recall the fact that the limit of the πth power of a function is equal to the limit of the function always to the πth power if π is a positive integer. Using this, we have that the limit as π₯ approaches β of the square of the function of one over π₯ is equal to the square of the limit as π₯ approaches β of the function one over π₯.

Now, recall the standard result that the limit as π₯ approaches β of the function one over π₯ is equal to zero. Using this, we find that the limit in question is equal to negative three over negative four plus eight times zero, which simplifies to negative three over negative four and further to three over four. So the limit as π₯ approaches β of the function negative three π₯ squared over negative four π₯ squared plus eight is equal to three over four.

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