AP Calculus AB and BC

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

Q: How do I compare degrees when a square root is involved?

Halve the degree of the radicand, and take the square root of its leading coefficient. Here \sqrt{x^2+1} counts as degree 1 with coefficient 1, so the limit is \frac{2}{1} = 2. By the same reading, \frac{2x+3}{\sqrt{9x^2+1}} has limit \frac{2}{3}.

Q: Why is the limit -2 at negative infinity?

The denominator stays positive at both ends, since square roots return nonnegative values, while the numerator 2x+3 is negative far to the left. A negative over a positive is negative, and the sizes still match, so the value settles at -2.

Q: Does L'Hopital's rule give the wrong answer here, or just no answer?

Just no answer. Every hypothesis holds and every step is valid, but the sequence alternates between the expression and a multiple of its reciprocal, so it never terminates. Dividing by the dominant term finishes in two lines.

The limit of (2x + 3) over the square root of x squared plus 1, as x approaches infinity, is 2. Substitution gives infinity over infinity. The radical grows like the absolute value of x, which equals x on the positive side, so dividing by x leaves the ratio 2 over 1. At negative infinity the same work gives -2.

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

Settled by dividing by the dominant term.

Pulling x out of the radical

Heading to $+\infty$ lets us assume $x > 0$ , and on that side $\sqrt{x^2} = x$ . Factor $x^2$ out from under the root and it comes out as a plain $x$ .

\sqrt{x^2+1} = \sqrt{x^2\left(1+\frac{1}{x^2}\right)} = x\sqrt{1+\frac{1}{x^2}} \qquad (x > 0)

Dividing the numerator by the same $x$ turns the whole fraction into a comparison of constants.

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

Now $\frac{3}{x}$ and $\frac{1}{x^2}$ both go to $0$ , the radicand goes to $1$ , and the root of $1$ is $1$ .

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

$x$	$\frac{2x+3}{\sqrt{x^2+1}}$
$10$	$2.28859$
$100$	$2.02990$
$1000$	$2.00300$

The $+3$ is what keeps those values above $2$ , and it loses its grip at the rate $\frac{3}{x}$ , which is why the convergence is slow enough to see.

Why substitution fails, and why L'Hopital stalls

Both parts are unbounded, so substitution returns $\frac{\infty}{\infty}$ and decides nothing. What makes this one look harder than a rational function is that the radical hides the degree.

A square root halves growth. The radicand $x^2+1$ has degree $2$ , so $\sqrt{x^2+1}$ behaves like degree $1$ , matching the numerator. Equal growth is exactly the situation that produces a finite nonzero answer.

L'Hopital's rule is legal here and gets nowhere. The numerator differentiates to $2$ , the denominator to $\frac{x}{\sqrt{x^2+1}}$ by the chain rule, and the new quotient is the same shape turned upside down.

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

Another pass sends it back again. Verifying the indeterminate form buys permission to use the rule, not a promise that the next line is simpler. When the derivative of a radical rebuilds the original expression, switch to dividing by the dominant term.

The other end changes the sign

For $x < 0$ the factoring step carries a minus, because $\sqrt{x^2} = |x| = -x$ on that side. A square root never returns a negative number, so the $x$ coming out from under the root has to be corrected.

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

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

A sign check is faster than the algebra. Far to the left the numerator $2x+3$ is negative while the denominator is a square root and therefore positive, so the quotient cannot come out positive. At $x = -1000$ the value is about $-1.99700$ .

So the graph has two horizontal asymptotes, $y = 2$ and $y = -2$ . Any time a variable moves in or out of a square root, check the sign before trusting the step.

The mistakes students make

Writing $\sqrt{x^2+1} = x+1$ . Roots do not open across a sum: at $x = 3$ the left side is about $3.162$ and the right side is $4$ .
Writing $\sqrt{x^2} = x$ with no bars. It is invisible heading to $+\infty$ and it flips the answer heading to $-\infty$ , turning $-2$ into $2$ .
Answering $\infty$ because the top looks like degree $1$ and the bottom like degree $2$ . The root halves the degree of the radicand, so both are degree $1$ .
Keeping the $3$ in the answer. It divides down to $\frac{3}{x}$ and disappears, affecting how the graph approaches rather than what it approaches.
Grinding L'Hopital's rule until something changes. It cycles here, and cycling is a signal to switch methods.

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

How do I compare degrees when a square root is involved?

Halve the degree of the radicand, and take the square root of its leading coefficient. Here $\sqrt{x^2+1}$ counts as degree $1$ with coefficient $1$ , so the limit is $\frac{2}{1} = 2$ . By the same reading, $\frac{2x+3}{\sqrt{9x^2+1}}$ has limit $\frac{2}{3}$ .

Why is the limit -2 at negative infinity?

The denominator stays positive at both ends, since square roots return nonnegative values, while the numerator $2x+3$ is negative far to the left. A negative over a positive is negative, and the sizes still match, so the value settles at $-2$ .

Does L'Hopital's rule give the wrong answer here, or just no answer?

Just no answer. Every hypothesis holds and every step is valid, but the sequence alternates between the expression and a multiple of its reciprocal, so it never terminates. Dividing by the dominant term finishes in two lines.