AP Calculus AB and BC

Limit of sqrt(9x^2+1)/x as x Approaches Infinity

Q: Why is the limit negative at negative infinity?

Because \sqrt{9x^{2}+1} is always positive while x is negative, so the quotient is negative. Formally \sqrt{x^{2}} = |x|, which is -x when x is negative.

The limit of the square root of nine x squared plus one, divided by x, as x approaches infinity is three. The root halves the degree inside, so the numerator behaves like three x and the quotient settles at three.

\lim_{x \to \infty} \frac{\sqrt{9x^{2}+1}}{x} = 3

Settled by dividing by the dominant term.

Move the x inside the root

For $x > 0$ , dividing by $x$ is the same as dividing inside the root by $x^{2}$ .

\frac{\sqrt{9x^{2}+1}}{x} = \sqrt{\frac{9x^{2}+1}{x^{2}}} = \sqrt{9 + \frac{1}{x^{2}}} \longrightarrow \sqrt{9} = 3

So the line $y = 3$ is a horizontal asymptote on the right. The constant 1 inside the root contributes nothing in the limit.

The sign trap going the other way

The step above used $\sqrt{x^{2}} = x$ , which is only true for POSITIVE $x$ . As $x \to -\infty$ the correct identity is $\sqrt{x^{2}} = |x| = -x$ , and the limit becomes $-3$ .

So this function has TWO horizontal asymptotes, $y = 3$ and $y = -3$ . Any square root of an even power at negative infinity needs this check; it is the single most common error in end-behaviour problems.

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

Why is the limit negative at negative infinity?

Because $\sqrt{9x^{2}+1}$ is always positive while $x$ is negative, so the quotient is negative. Formally $\sqrt{x^{2}} = |x|$ , which is $-x$ when x is negative.

How do I handle a root of an even power at infinity?

Factor out the highest even power, take the root as an absolute value, then resolve the absolute value using the sign of x on the side you are approaching from.