AP Calculus AB and BC

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

Q: Why does the answer flip sign at negative infinity?

A square root never returns a negative output, so \sqrt{x^{2}+1} stays positive on both ends. The numerator x does not: it is negative on the left. A negative over a positive is negative, and the size comparison is unchanged, so the value approaches -1.

The limit of x over the square root of x squared plus 1, as x approaches infinity, is 1. Divide top and bottom by x, noting that moving x inside the radical turns it into the square root of x squared, which is the absolute value of x and equals x only when x is positive. At negative infinity the same work gives -1.

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

Settled by dividing by the dominant term.

Why x is the term that matters

For large $x$ the radical is doing almost exactly what $x$ is doing, since the $+1$ under the root is swamped by $x^{2}$ . Numerator and denominator therefore grow at the same rate, and once that shared rate is stripped out of both, a constant is what remains.

Heading to $\infty$ lets us take $x > 0$ , and on that side $x = \sqrt{x^{2}}$ . So dividing the denominator by $x$ is the same as pulling $x^{2}$ under the radical, where it combines with what is already there.

\frac{\sqrt{x^{2}+1}}{x} = \frac{\sqrt{x^{2}+1}}{\sqrt{x^{2}}} = \sqrt{\frac{x^{2}+1}{x^{2}}} = \sqrt{1 + \frac{1}{x^{2}}} \quad (x > 0)

The numerator divided by $x$ is simply 1, so the whole expression collapses to something with no competition left in it.

\frac{x}{\sqrt{x^{2}+1}} = \frac{1}{\sqrt{1 + \frac{1}{x^{2}}}} \quad (x > 0)

Now $\frac{1}{x^{2}}$ goes to 0, the radicand goes to 1, and the square root of 1 is 1.

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

$x$	$\frac{x}{\sqrt{x^{2}+1}}$
$10$	$0.99504$
$100$	$0.99995$
$1000$	$0.9999995$

The values climb toward 1 from below and never reach it, since $\sqrt{x^{2}+1}$ is always strictly larger than $x$ .

Why substitution fails, and why L'Hopital stalls

Both parts run to $\infty$ , so substitution returns $\frac{\infty}{\infty}$ , which decides nothing on its own. The whole content of the problem is which part wins, and the form records only that both are large.

L'Hopital's rule is legal here, since the form is one of the two it accepts. It is also useless. The numerator differentiates to 1, the denominator differentiates to $\frac{x}{\sqrt{x^{2}+1}}$ by the chain rule, and the new quotient is the reciprocal of the original.

\lim_{x \to \infty} \frac{x}{\sqrt{x^{2}+1}} \;\longrightarrow\; \lim_{x \to \infty} \frac{1}{\frac{x}{\sqrt{x^{2}+1}}} = \lim_{x \to \infty} \frac{\sqrt{x^{2}+1}}{x}

Apply it once more and the original expression comes back. The rule cycles between a function and its reciprocal forever, so no amount of persistence produces an answer.

Legal and useful are different tests

Verifying the indeterminate form only earns you permission to apply the rule. It does not promise the new limit is easier. When the derivative of the denominator rebuilds the original expression, as it does with radicals of this shape, stop and divide by the dominant term instead.

The absolute value trap

The step that gets written down wrong is $\sqrt{x^{2}} = x$ . A square root returns the nonnegative root, so the correct identity carries bars.

\sqrt{x^{2}} = |x| = \begin{cases} x & x \ge 0 \\ -x & x < 0 \end{cases}

Approaching $+\infty$ the error is invisible, because $x$ is positive and the two agree. Approaching $-\infty$ it changes the answer. Dividing by a negative $x$ means $\frac{\sqrt{x^{2}+1}}{x} = \frac{\sqrt{x^{2}+1}}{-|x|}$ , so a minus sign survives.

\frac{x}{\sqrt{x^{2}+1}} = \frac{-1}{\sqrt{1 + \frac{1}{x^{2}}}} \quad (x < 0)

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

A sanity check beats the algebra anyway: for negative $x$ the numerator is negative while the denominator is a square root and therefore positive, so the quotient cannot come out positive.

The graph has two horizontal asymptotes, $y = 1$ and $y = -1$ . Any time a variable moves in or out of a square root, check the sign before trusting the cancellation.

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 does the answer flip sign at negative infinity?

A square root never returns a negative output, so $\sqrt{x^{2}+1}$ stays positive on both ends. The numerator $x$ does not: it is negative on the left. A negative over a positive is negative, and the size comparison is unchanged, so the value approaches $-1$ .

Can I just compare leading terms the way I would for a rational function?

The instinct is right and the bookkeeping is riskier. The dominant term inside the radical is $x^{2}$ , whose root is $|x|$ rather than $x$ , so the shortcut reads the function as $\frac{x}{|x|}$ . That gives 1 on the right and $-1$ on the left, which is correct, but only if you keep the bars.

Is L'Hopital's rule ever wrong here, or just unhelpful?

Just unhelpful. The hypotheses hold, so every step is valid; the sequence simply never terminates, alternating between the expression and its reciprocal. A cycling L'Hopital is a signal to switch techniques, not a sign that you made an arithmetic error.