AP Calculus AB and BC

Limit of (3x^2 + 2x) / (x^2 - 5) at Infinity

Q: What is the limit of (3x^2 + 2x) / (x^2 - 5) as x approaches infinity?

It is 3, the ratio of the leading coefficients.

Q: Do the 2x and -5 matter at all?

Not to the limit, but they decide how the graph approaches y = 3. Here it approaches from above for large positive x.

Q: Can I use L'Hopital's rule?

Yes, the form is \frac{\infty}{\infty}. Two passes give \frac{6}{2} = 3.

The limit of 3x squared plus 2x, over x squared minus 5, as x approaches infinity is 3. Both parts have degree 2, so the limit is the ratio of the leading coefficients, 3 over 1. The lower-order terms 2x and minus 5 affect the approach but not the destination.

\lim_{x \to \infty} \frac{3x^{2} + 2x}{x^{2} - 5} = 3

Settled by dividing by the highest power.

Dividing by x squared

Divide every term by $x^{2}$ , the highest power anywhere in the expression.

\frac{3x^{2}+2x}{x^{2}-5} = \frac{3 + \frac{2}{x}}{1 - \frac{5}{x^{2}}}

Every correction term vanishes, and the denominator's limit is $1$ , safely nonzero, so the quotient law applies.

\lim_{x \to \infty}\frac{3+\frac{2}{x}}{1-\frac{5}{x^{2}}} = \frac{3+0}{1-0} = 3

What the lower-order terms actually do

They control HOW the function approaches $3$ , not what it approaches. The $+2x$ on top keeps the values above $3$ for large positive $x$ : at $x = 100$ the value is about $3.021$ .

This is worth saying because a graph question may ask whether the curve approaches its asymptote from above or below, and the leading coefficients alone cannot answer that.

L'Hopital works but is slower

The form is infinity over infinity, so the rule is legal. Two passes give 6 over 2, which is 3. Dividing by the highest power is faster and shows the approach direction as a bonus.

The mistakes students make

Answering $\infty$ because both parts grow. Equal degrees give a finite ratio.
Including the lower-order terms in the answer, as in $3 + 2$ . They vanish once divided by $x^{2}$ .
Cancelling $x^{2}$ term by term out of a sum. Only a common FACTOR can be cancelled, and $x^{2}$ is not a factor of $3x^{2}+2x$ once you also need it to divide the $-5$ .

Not sure which technique a limit wants?

The Limit Method Chooser walks the decision from direct substitution through factoring, the conjugate, and L'Hopital, and says why each one applies or fails.

Frequently asked questions

What is the limit of (3x^2 + 2x) / (x^2 - 5) as x approaches infinity?

It is $3$ , the ratio of the leading coefficients.

Do the 2x and -5 matter at all?

Not to the limit, but they decide how the graph approaches $y = 3$ . Here it approaches from above for large positive $x$ .

Can I use L'Hopital's rule?

Yes, the form is $\frac{\infty}{\infty}$ . Two passes give $\frac{6}{2} = 3$ .