AP Calculus AB and BC

Limit of 1/x as x Approaches 0 from the Right

Q: Should I write infinity or DNE?

Write \infty. It says everything a bare non-existence claim says and adds the direction, which is what the asymptote question behind it usually needs. Save the plain non-existence answer for cases like \lim_{x \to 0} \frac{|x|}{x}, where the two sides settle on different finite numbers.

The limit of 1 over x as x approaches 0 from the right is infinity. The denominator shrinks through positive values, so the quotient grows past any bound you name. Approaching from the left gives negative infinity instead, which is why the two-sided limit at 0 does not exist.

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

Settled by one-sided unbounded behaviour.

Marching in from the right

Every $x$ to the right of 0 is positive, so $\frac{1}{x}$ is positive as well. Shrinking $x$ toward 0 shrinks the denominator without ever letting it arrive, and a fixed numerator over a vanishing positive denominator has to grow.

$x$	$\frac{1}{x}$
$0.1$	$10$
$0.01$	$100$
$10^{-3}$	$10^{3}$
$10^{-6}$	$10^{6}$

What the table shows is not merely large values, it is values with no ceiling. Name any bound $M$ , however big. Every $x$ with $0 < x < \frac{1}{M}$ satisfies $\frac{1}{x} > M$ , and the function stays past that bound for the rest of the approach.

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

That statement records unbounded growth. No real number is being approached, so the limit does not exist as a number, and $\infty$ is the notation for how it fails.

Why substitution fails, and why 1 over 0 is not indeterminate

Substitution gives $\frac{1}{0}$ , which is undefined. Students often file that alongside $\frac{0}{0}$ , but the two behave nothing alike. $\frac{0}{0}$ is indeterminate because a shrinking numerator and a shrinking denominator compete, and which one wins depends on the functions. Here the numerator sits at 1 and competes with nothing.

So the size of the answer is settled before any work starts: a fixed nonzero numerator over a denominator collapsing to 0 always produces unbounded values. The only open question is the sign, and on the right of 0 the denominator is positive, so the values run up.

L'Hopital's rule has no license here

The rule needs a confirmed $\frac{0}{0}$ or $\frac{\infty}{\infty}$ . Differentiating top and bottom of $\frac{1}{x}$ would give $\frac{0}{1} = 0$ , a wrong answer produced by applying a rule to a form it was never meant for.

The geometry of the same fact is the vertical asymptote at $x = 0$ . That pairing is Topic 1.14, connecting infinite limits and vertical asymptotes, which puts this in Unit 1.

The mistake students make

Dropping the superscript and writing $\lim_{x \to 0} \frac{1}{x} = \infty$ . The left side runs to $-\infty$ , so the two-sided limit does not exist, not even as an infinite limit.
Answering 0, out of habit from $\lim_{x \to \infty} \frac{1}{x} = 0$ . That is the reciprocal situation: a huge denominator gives a tiny value, a tiny denominator gives a huge one.
Treating $\frac{1}{0}$ as indeterminate and hunting for algebra. There is nothing to cancel and no competition to resolve.
Answering that the limit is undefined and stopping. The rubric wants the direction, so $\infty$ is the expected form.

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 two-sided limit not exist?

The two one-sided limits disagree. From the right the denominator is positive and $\frac{1}{x} \to \infty$ ; from the left it is negative and $\frac{1}{x} \to -\infty$ . A two-sided limit needs both sides doing the same thing, including when that thing is running off to infinity.

Should I write infinity or DNE?

Write $\infty$ . It says everything a bare non-existence claim says and adds the direction, which is what the asymptote question behind it usually needs. Save the plain non-existence answer for cases like $\lim_{x \to 0} \frac{|x|}{x}$ , where the two sides settle on different finite numbers.

How is this different from 1 over x squared?

Squaring makes the denominator positive on both sides, so both one-sided limits are $\infty$ and the two-sided limit is $\infty$ too. The odd power in $\frac{1}{x}$ is the only thing splitting the two sides apart.