AP Calculus AB and BC

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

Q: Does the limit change as x approaches negative infinity?

No, it is \frac{1}{2} at both ends. The dominant terms x^2 and 2x^2 are even powers, so their ratio ignores the sign of x, and y = \frac{1}{2} is a single horizontal asymptote for the whole graph.

Q: What if the denominator were 2x^3 - 3 instead?

Then the limit would be 0. The denominator degree would exceed the numerator degree, so dividing by x^3 leaves \frac{\frac{1}{x} + \frac{1}{x^3}}{2 - \frac{3}{x^3}}, whose top goes to 0 while the bottom goes to 2. The horizontal asymptote would be y = 0.

The limit of (x^2 + 1)/(2x^2 - 3) as x approaches infinity is 1/2. Substitution gives infinity over infinity. Divide top and bottom by x squared and the leftovers vanish, leaving the ratio of the leading coefficients, 1 over 2. The line y = 1/2 is the horizontal asymptote at both ends.

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

Settled by comparing leading degrees.

Dividing by x squared

Divide every term on the top and the bottom by $x^2$ , the highest power present. Both parts get the same treatment, so the fraction is the same fraction afterwards.

\frac{x^2+1}{2x^2-3} = \frac{1 + \frac{1}{x^2}}{2 - \frac{3}{x^2}}

Both $\frac{1}{x^2}$ and $\frac{3}{x^2}$ go to $0$ , and what is left is a quotient of constants.

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

The approach is from above: the $+1$ inflates the top while the $-3$ shrinks the bottom, so every value sits a little over $\frac{1}{2}$ .

$x$	$\frac{x^2+1}{2x^2-3}$
$10$	$0.512690$
$100$	$0.500125$
$1000$	$0.5000013$

Why substitution fails

Substituting sends both parts to $\infty$ and produces $\frac{\infty}{\infty}$ , which is indeterminate. Two unbounded quantities can settle at any ratio at all, and the form keeps no record of their speeds.

At $x = 1000$ the numerator is $1{,}000{,}001$ and the denominator is $1{,}999{,}997$ . Neither the $+1$ nor the $-3$ is visible at that scale, which is the whole content of the degree rule: at the far end a polynomial is its leading term.

Which coefficients, and in which order

Equal degrees means the limit is the ratio of leading coefficients. The numerator's leading coefficient is the unwritten $1$ in front of $x^2$ , and the denominator's is $2$ .

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

Order matters, and $\frac{1}{2}$ is not $2$ . The coefficient that started on top stays on top. A quick sanity check settles it in one line: the denominator is roughly twice the numerator for large $x$ , so the fraction has to be roughly $\frac{1}{2}$ , not $2$ .

The dominant powers are even, so the sign of $x$ never enters and the same value holds in the other direction.

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

The graph never meets its asymptote

Setting $\frac{x^2+1}{2x^2-3} = \frac{1}{2}$ gives $2x^2+2 = 2x^2-3$ , that is $2 = -3$ , so there is no solution and the curve never crosses $y = \frac{1}{2}$ . Crossing is legal in general, though: plenty of rational functions cut through their horizontal asymptote before settling onto it. This one just happens not to.

The mistakes students make

Answering $2$ by flipping the ratio. The $2$ belongs to the denominator, so it stays underneath.
Missing the invisible $1$ . The numerator $x^2+1$ has leading coefficient $1$ , and the $+1$ at the end is a constant term, not a coefficient of $x^2$ .
Cancelling the $x^2$ terms straight across. It happens to land on $\frac{1}{2}$ here and stops working the moment the degrees differ.
Answering $0$ because the numerator looks smaller. Smaller by a constant factor is not smaller in degree, and only degree decides whether the limit is $0$ .
Worrying about $x = \pm\sqrt{\frac{3}{2}} \approx \pm 1.2247$ , where the denominator vanishes. Those are vertical asymptotes at finite inputs and have nothing to say about behaviour at infinity.

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 answer 1/2 rather than 2?

The ratio keeps the original orientation: numerator coefficient over denominator coefficient, so $\frac{1}{2}$ . For large $x$ the denominator is about twice the numerator, so the fraction must be about one half. Checking one value settles it, since at $x = 10$ the function is $\frac{101}{197} \approx 0.513$ .

Does the limit change as x approaches negative infinity?

No, it is $\frac{1}{2}$ at both ends. The dominant terms $x^2$ and $2x^2$ are even powers, so their ratio ignores the sign of $x$ , and $y = \frac{1}{2}$ is a single horizontal asymptote for the whole graph.

What if the denominator were 2x^3 - 3 instead?

Then the limit would be $0$ . The denominator degree would exceed the numerator degree, so dividing by $x^3$ leaves $\frac{\frac{1}{x} + \frac{1}{x^3}}{2 - \frac{3}{x^3}}$ , whose top goes to $0$ while the bottom goes to $2$ . The horizontal asymptote would be $y = 0$ .