AP Calculus AB and BC

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

Q: Is this just the reciprocal of the limit of x/sqrt(x^2+1)?

Yes. That limit is 1, and reciprocals invert limits whenever the limit is not 0, so this one is \frac{1}{1} = 1. Had the first limit been 0, the reciprocal rule would not apply and the answer would be an unbounded one instead.

Q: Why do the values stay above 1?

Because \sqrt{x^2+1} is strictly larger than \sqrt{x^2} = x for positive x, so the numerator always beats the denominator by a shrinking margin. At x = 10 that margin puts the function at about 1.004988, and the gap closes like \frac{1}{2x^2}.

Q: What about sqrt(4x^2+1)/x or sqrt(x^2+x)/x?

They give 2 and 1. Factoring out x^2 turns the first into \sqrt{4+\frac{1}{x^2}} \to 2 and the second into \sqrt{1+\frac{1}{x}} \to 1. The leading coefficient inside the radical comes out as its square root, which is the rule worth remembering.

The limit of the square root of x squared plus 1, divided by x, as x approaches infinity, is 1. Substitution gives infinity over infinity. For positive x it equals the square root of 1 plus 1 over x squared, which tends to 1. At negative infinity it is -1, since the square root of x squared is the absolute value of x.

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

Settled by dividing by the dominant term.

Moving x under the root

Heading to $+\infty$ means $x > 0$ , and on that side $x = \sqrt{x^2}$ . So dividing by $x$ is the same as dividing by $\sqrt{x^2}$ , which slides straight under the radical and 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}} \qquad (x > 0)

Nothing indeterminate is left. The $\frac{1}{x^2}$ goes to $0$ , the radicand goes to $1$ , and the square root is continuous at $1$ .

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

$x$	$\frac{\sqrt{x^2+1}}{x}$
$10$	$1.004988$
$100$	$1.0000500$
$1000$	$1.0000005$

Every value sits above $1$ , since $x^2+1 > x^2$ forces $\sqrt{x^2+1} > x$ for positive $x$ . The graph settles onto $y = 1$ from above and never touches it.

Why substitution fails, and why L'Hopital cycles

Both parts run to $\infty$ , so substitution returns $\frac{\infty}{\infty}$ , which is silent about the ratio. The tempting repair is a degree count: top of degree $2$ , bottom of degree $1$ , answer $\infty$ . That count is wrong because a square root halves growth, so $\sqrt{x^2+1}$ behaves like degree $1$ and matches the denominator.

L'Hopital's rule is available, since the form is one it accepts, and it goes in circles. Differentiating the radical by the chain rule gives $\frac{x}{\sqrt{x^2+1}}$ , which is the reciprocal of the original expression.

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

One more pass brings the original back, so the rule alternates forever between a function and its reciprocal. Legal and useful are separate tests, and a cycling L'Hopital means switch methods rather than keep going.

The absolute value at the far left

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

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

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

A sign check confirms it without algebra: the numerator is a square root and therefore positive, the denominator is negative far to the left, and a positive over a negative is negative. At $x = -1000$ the value is about $-1.0000005$ .

So the graph has two horizontal asymptotes, $y = 1$ and $y = -1$ , and it approaches each from outside.

The mistakes students make

Writing $\sqrt{x^2+1} = x+1$ . At $x = 3$ that claims $4$ where the true value is about $3.162$ ; roots do not open across a sum.
Writing $\sqrt{x^2} = x$ with no bars, which silently reports $1$ at $-\infty$ where the answer is $-1$ .
Answering $\infty$ from a degree count that ignores the root. Halve the degree of the radicand before comparing.
Cancelling the $x^2$ under the radical against the $x$ outside it. The outside $x$ has to be squared before it can enter the root, which is exactly what $\sqrt{x^2}$ records.
Reporting that the limit does not exist because the two ends disagree. Each end has its own limit, $1$ and $-1$ , and both exist.

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

Is this just the reciprocal of the limit of x/sqrt(x^2+1)?

Yes. That limit is $1$ , and reciprocals invert limits whenever the limit is not $0$ , so this one is $\frac{1}{1} = 1$ . Had the first limit been $0$ , the reciprocal rule would not apply and the answer would be an unbounded one instead.

Why do the values stay above 1?

Because $\sqrt{x^2+1}$ is strictly larger than $\sqrt{x^2} = x$ for positive $x$ , so the numerator always beats the denominator by a shrinking margin. At $x = 10$ that margin puts the function at about $1.004988$ , and the gap closes like $\frac{1}{2x^2}$ .

What about sqrt(4x^2+1)/x or sqrt(x^2+x)/x?

They give $2$ and $1$ . Factoring out $x^2$ turns the first into $\sqrt{4+\frac{1}{x^2}} \to 2$ and the second into $\sqrt{1+\frac{1}{x}} \to 1$ . The leading coefficient inside the radical comes out as its square root, which is the rule worth remembering.