Video Transcript
Find the horizontal asymptotes of the function π of π₯ is equal to three π₯ squared plus seven divided by five π₯ squared minus four.
The question wants us to find the horizontal asymptotes of our rational function π of π₯. And we recall, we say that the line π¦ is equal to π is a horizontal asymptote to the curve π¦ is equal to π of π₯ if the limit as π₯ approaches β of π of π₯ is equal to π or the limit as π₯ approaches negative β of π of π₯ is equal to π. So to find the possible horizontal asymptotes of our function, we just need to calculate the limit as π₯ approaches β and the limit as π₯ approaches negative β of our rational function π of π₯.
Letβs start by calculating the limit as π₯ approaches β of π of π₯. Thatβs the limit as π₯ approaches β of three π₯ squared plus seven divided by five π₯ squared minus four. If we tried to evaluate this limit directly, we would see that we would get β divided by β. This is an indeterminate form. So weβll have to evaluate this limit in a different way.
We can actually do this by dividing both our numerator and our denominator by the highest power of π₯ which appears in the fraction. We see that the highest power of π₯ which appears in our fraction is π₯ squared. So weβre going to divide both our numerator and our denominator through by π₯ squared.
In our numerator, we have three π₯ squared divided by π₯ squared is just three. And seven divided by π₯ squared is seven over π₯ squared. In our denominator, we have five π₯ squared divided by π₯ squared is five. And then negative four divided by π₯ squared is negative four divided by π₯ squared. And we can now see by evaluating this limit since π₯ is approaching β, seven divided by π₯ squared and negative four divided by π₯ squared are both approaching zero.
And we know that our constants of three and five are not changing with π₯. So we can actually just evaluate this limit to give us three divided by five. We could then write out and evaluate the limit as π₯ approaches negative β of π of π₯ in a similar way. However, we could also ask the question, what wouldβve happened if, instead of weβd taken the limit as π₯ approaches β, weβd taken the limit as π₯ approaches negative β?
If weβd instead taken the limit as π₯ approaches negative β, we can still rewrite π of π₯ as three π₯ squared plus seven divided by five π₯ squared minus four. And we can still divide through our numerator and our denominator by π₯ squared. And dividing through by π₯ squared does not change.
We now need to calculate the limit as π₯ approaches negative β of three divided by seven over π₯ squared all divided by five minus four over π₯ squared. And we can see the same line of reasoning holds. Seven over π₯ squared and negative four over π₯ squared still both approach zero as π₯ approaches negative β. And three and five are still both constants even when π₯ is approaching negative β. So the limit as π₯ approaches negative β of π of π₯ is also equal to three divided by five.
Therefore, weβve shown that the only horizontal asymptote of our function π of π₯ is equal to three π₯ squared plus seven all divided by five π₯ squared minus four is the line π¦ is equal to three-fifths.