AP Calculus AB and BC

Limit of (3x-2)/sqrt(2x^2+1) at Infinity Is 3/sqrt2

Q: What is the limit of (3x-2)/sqrt(2x^2+1) as x approaches infinity?

It is \frac{3\sqrt{2}}{2}, which is about 2.121.

Q: Why is the answer irrational?

Because the dominant part of the denominator is \sqrt{2x^{2}} = \sqrt{2}\,|x|. The \sqrt{2} survives into the answer, which a rational function with whole-number coefficients never produces.

Q: What happens at negative infinity?

The limit is -\frac{3\sqrt{2}}{2}, since \sqrt{2x^{2}+1} behaves like \sqrt{2}\,|x| = -\sqrt{2}\,x for negative x.

The limit of 3x minus 2, over the square root of 2x squared plus 1, as x approaches infinity is 3 over the square root of 2, which rationalises to 3 root 2 over 2, about 2.121. Dividing by x sends that x inside the root as x squared, leaving 3 over root 2.

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

Settled by dividing by the dominant term.

Divide by x, and by x squared inside the root

Under the radical the term $2x^{2}$ counts as degree one, because the root halves the degree. Top and bottom therefore grow at the same rate and the quotient has a finite limit.

\frac{3x-2}{\sqrt{2x^{2}+1}} = \frac{3 - \dfrac{2}{x}}{\sqrt{2 + \dfrac{1}{x^{2}}}} \qquad (x > 0)

\lim_{x \to \infty}\frac{3x-2}{\sqrt{2x^{2}+1}} = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2} \approx 2.121

An asymptote that is not a ratio of integers

When a rational function has the same degree top and bottom, the horizontal asymptote is the ratio of the two leading coefficients, so with whole-number coefficients it comes out rational. A radical breaks that pattern: here the denominator contributes $\sqrt{2}$ rather than $2$ , and the asymptote is irrational.

On the left the same reasoning with $\sqrt{x^{2}} = -x$ gives $-\frac{3\sqrt{2}}{2}$ , so the graph has one asymptote at each end and they are reflections of one another.

A quick sanity check

Put x = 1000. The numerator is 2998 and the denominator is about 1414.2, giving roughly 2.120. That matches 3 root 2 over 2 and rules out any answer near 1.5.

The mistakes students make

All three follow from mishandling the radical: its degree, its coefficient, or the sign it carries at the negative end.

Answering $\frac{3}{2}$ by taking the leading coefficients $3$ and $2$ straight, without square rooting the $2$ .
Answering $0$ because the denominator contains $x^{2}$ . Under a square root that term behaves like $x$ , matching the numerator's degree.
Answering $+\frac{3\sqrt{2}}{2}$ at negative infinity as well. Since $\sqrt{2x^{2}+1}$ behaves like $\sqrt{2}\,|x| = -\sqrt{2}\,x$ for negative $x$ , the left-hand limit is $-\frac{3\sqrt{2}}{2}$ .

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)/sqrt(2x^2+1) as x approaches infinity?

It is $\frac{3\sqrt{2}}{2}$ , which is about $2.121$ .

Why is the answer irrational?

Because the dominant part of the denominator is $\sqrt{2x^{2}} = \sqrt{2}\,|x|$ . The $\sqrt{2}$ survives into the answer, which a rational function with whole-number coefficients never produces.

What happens at negative infinity?

The limit is $-\frac{3\sqrt{2}}{2}$ , since $\sqrt{2x^{2}+1}$ behaves like $\sqrt{2}\,|x| = -\sqrt{2}\,x$ for negative $x$ .