Video: APCALC04AB-P1A-Q08-319143784942

Find lim_(π‘₯ β†’ ∞) (7π‘₯Β² + 5π‘₯ βˆ’ 8)/(9π‘₯Β² βˆ’ π‘₯ + 6).

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Video Transcript

Find the limit as π‘₯ approaches ∞ of seven π‘₯ squared plus five π‘₯ minus eight over nine π‘₯ squared minus π‘₯ plus six.

Here, we have a rather nasty-looking limit. It’s the quotient of two quadratic expressions. And we’re looking to evaluate this when π‘₯ approaches ∞. We cannot go ahead and use direct substitution. When we do, we obtain ∞ divided by ∞, which we know to be undefined. So we alternatively look for a way to manipulate our expression.

We have two quadratics. So we might try to factor the numerator and the denominator here. However, if we do, we see that neither expression is indeed factorable. In fact, the discriminant of nine π‘₯ squared minus π‘₯ plus six is less than zero. That tells us that the equation nine π‘₯ squared minus π‘₯ plus six equals zero has no roots. And indeed, nine π‘₯ squared minus π‘₯ plus six is not factorable. So instead, we identify the largest power of π‘₯ in our denominator. That’s π‘₯ squared. And then, we divide everything on the numerator and the denominator of our fraction by π‘₯ squared.

By simplifying, we find that we need to evaluate the limit as π‘₯ approaches ∞ of seven minus five over π‘₯ minus eight over π‘₯ squared over nine minus one over π‘₯ plus six over π‘₯ squared. And actually, now, we’re ready to use direct substitution. As π‘₯ approaches ∞, five over π‘₯, eight over π‘₯ squared, one over π‘₯, and six over π‘₯ squared approach zero. And that leaves us with seven minus zero minus zero over nine minus zero plus zero, which is seven-ninths.

And so the limit as π‘₯ approaches ∞ of seven π‘₯ squared plus five π‘₯ minus eight over nine π‘₯ squared minus π‘₯ plus six is seven-ninths.

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