# Video: Finding the Limit at Infinity of a Rational Function given the Degree of the Dividend and the Divisor Polynomials

If π(π₯) is a polynomial function of a fifth degrees, and π(π₯) is a polynomial function of a fourth degrees, find lim_(π₯ β β) (π(π₯))/(4π₯β΅ π(π₯)).

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### Video Transcript

If π of π₯ is a polynomial function of fifth degree and π of π₯ is a polynomial function of fourth degree. Find the limit as π₯ tends to β of π of π₯ over four π₯ to the power of five multiplied by π of π₯.

Since π of π₯ is a polynomial function of fifth degree, let π of π₯ equal π five timesed by π₯ to the power of five. Plus π four timesed by π₯ to the power of four. Plus π three timesed by π₯ to the power of three. Plus π two timesed by π₯ to the power of two. Plus π one π₯ plus π zero. Where π zero up to π five are constants and π five is nonzero.

Since π of π₯ is a polynomial function of fourth degree, let π of π₯ equal π four timesed by π₯ to the power of four. Plus π three timesed by π₯ to the power of three. Plus π two timesed by π₯ to the power of two. Plus π one π₯ plus π zero. Where π zero up to π four are constants and π four is nonzero.

We can substitute these expressions for π of π₯ and π of π₯ into the limit we are asked to evaluate, like so. Now letβs distribute the term four π₯ to the power of five over the parentheses in the denominator of the limit. Doing so, we obtain four π five π₯ to the power of 10. Plus four π four π₯ to the power of nine. Plus four π three π₯ to the power of eight. Plus four π two π₯ to the power of seven. Plus four π one π₯ to the power of six. Plus four π zero π₯ to the power of five.

At first, we might be inclined to use the result that the limit of a quotient of functions is equal to the quotient of their limits. Doing so, we would obtain the following. As π₯ tends to β, the polynomial expressions in the numerator and denominator just get larger and larger. And so we would obtain the indeterminate form β over β, which does not tell us the value of our limit.

So in order to evaluate our limit, we are going to use a different method. We are going to divide the numerator and denominator by the highest power of π₯ that is found in the denominator. This is π₯ to the power of 10. Doing so, we obtain this slightly complicated expression.

Letβs simplify this expression by simplifying the powers of π₯. 10 minus four is six. So this term simplifies to π four over π₯ to the power of six. 10 minus three is seven. So this term simplifies to π three over π₯ to the power of seven. 10 minus two is eight. So this term simplifies to π two over π₯ to the power of eight.

Similarly, we can simplify the rest of the terms in the numerator and denominator. Doing so, we obtain this slightly less complicated expression. Now letβs apply the fact that the limit of a quotient of functions is the quotient of their limits. Now letβs apply the fact that the limit of a sum of functions is the sum of their limits.

Now recall that we can pull constants out of a limit. Hence, we can pull the constants π four, π three, π two, π one, π zero, four π four, four π three, four π two, four π one, and four π zero out from the limits that they are in. We will come back to the limit of the constant four π five as π₯ tends to β in a moment.

Pulling the constants out from the limits that they were in, we obtain an expression that looks like this. Now recall that the limit of a constant function is just equal to that constant function itself. So the limit of the constant four times π five as π₯ tends to β is just equal to four times π five. Recall also that the limit as π₯ tends to β of the function one over π₯ to the power of π, where π is strictly greater than zero, is equal to zero.

Hence, all the remaining limits in the numerator and denominator of our expression are equal to zero. Hence, the numerator is equal to zero and the denominator is equal to four times π five. So we obtain that the limit in question is equal to zero over four times π five.

Now note that the constant π five is nonzero. And so the expression zero over four times π five is defined and, in fact, equal to zero. So we obtain that the limit in question is equal to zero.