AP Calculus AB and BC

Limit of (3x+1)/(x^2+2) at Infinity

Q: What is the horizontal asymptote?

y = 0. The limit as x \to \infty is 0 and the limit as x \to -\infty is also 0, approached from below there, so the x-axis is the asymptote in both directions.

Q: Do I need L'Hopital here?

No. Dividing by x^2 settles it in one step. L'Hopital works once, giving \frac{3}{2x} \to 0, but you have to stop there, because that expression is no longer an indeterminate form.

Q: What if the degrees were the other way around?

\frac{x^2+2}{3x+1} has the larger degree on top, so it grows without bound and has no horizontal asymptote. It has a slant asymptote instead, because the degrees differ by exactly one.

The limit of (3x + 1)/(x^2 + 2) as x approaches infinity is 0. Substitution gives infinity over infinity, so divide the top and bottom by x squared, the highest power in the denominator. The bottom outgrows the top, the quotient collapses, and y = 0 is the horizontal asymptote.

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

Settled by comparing leading degrees.

Dividing by the highest power in the denominator

Use $x^2$ , the highest power in the denominator. That choice always finishes the job, because it turns the bottom into 1 plus terms that vanish, which leaves the numerator to decide the answer on its own.

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

Every term upstairs now has an $x$ underneath, so all of them go to 0, while the denominator settles at 1.

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

Dividing by x also works

Dividing top and bottom by $x$ instead reaches the same answer. It gives $\frac{3 + \frac{1}{x}}{x + \frac{2}{x}}$ , where the numerator settles at 3 while the denominator runs off, so the quotient goes to $\frac{3}{\infty} = 0$ , since a fixed nonzero numerator over an unbounded denominator is forced to 0. That route finishes. Dividing by the denominator's power is still the habit worth building, because it always leaves a fraction of ordinary numbers to read off.

Check the size

At $x = 100$ the value is about $0.0301$ , at $x = 10^{3}$ about $0.00300$ , and at $x = 10^{4}$ about $0.000300$ . It is shrinking like $\frac{3}{x}$ , which is exactly what the leading terms predict.

Why substitution fails

Substitution gives $\frac{\infty}{\infty}$ , which is indeterminate: both parts are unbounded and the form does not say which grows faster. Here $x^2$ outpaces $3x$ by a widening factor, so the fraction is driven to 0, but that conclusion comes from the degrees and not from the form.

A quick way to see the size: divide the leading terms alone and you get $\frac{3x}{x^2} = \frac{3}{x}$ , which goes to 0. The lower-order terms cannot overturn that verdict, which is what the full division above confirms.

The mistake students make

Reading off the ratio of leading coefficients and answering 3. That rule applies only when the degrees are equal. Here the denominator's degree is higher, so the leading terms give $\frac{3x}{x^2} = \frac{3}{x}$ , and the limit is 0.
Reporting no horizontal asymptote because the limit is 0. Zero is a perfectly good asymptote, the line $y = 0$ .
Applying L'Hopital twice. One round gives $\frac{3}{2x}$ , which is no longer indeterminate, so the rule stops applying; a second round is an illegal step even though it happens to land on 0 here.
Assuming the graph stays above its asymptote. This one crosses $y = 0$ at $x = -\frac{1}{3}$ , where the numerator vanishes. A horizontal asymptote constrains the far ends, not the middle.

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 horizontal asymptote?

$y = 0$ . The limit as $x \to \infty$ is 0 and the limit as $x \to -\infty$ is also 0, approached from below there, so the $x$ -axis is the asymptote in both directions.

Do I need L'Hopital here?

No. Dividing by $x^2$ settles it in one step. L'Hopital works once, giving $\frac{3}{2x} \to 0$ , but you have to stop there, because that expression is no longer an indeterminate form.

What if the degrees were the other way around?

$\frac{x^2+2}{3x+1}$ has the larger degree on top, so it grows without bound and has no horizontal asymptote. It has a slant asymptote instead, because the degrees differ by exactly one.