AP Calculus AB and BC

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

Q: Does approaching from the left always give a negative answer?

No. The side tells you which inputs to test, not which sign to write. \lim_{x \to 0^-} \frac{1}{x^2} = \infty, because the square is positive on both sides. Test an input from the correct side and read the sign of the denominator.

Q: Can I write DNE instead of negative infinity?

The limit does fail to exist as a real number, but -\infty is the expected answer because it names the behaviour. Free response scoring treats the direction as part of describing the vertical asymptote, so the bare non-existence claim gives up information.

Q: What does this say about the graph?

The line x = 0 is a vertical asymptote. The left branch of the hyperbola plunges down along it while the right branch climbs, and the two branches never meet. That picture is the two-sided limit failing, drawn out.

The limit of 1 over x as x approaches 0 from the left is negative infinity. Every input on that side is negative, so the quotient is negative and its size grows past any bound. From the right the same function runs to positive infinity, so the two-sided limit at 0 does not exist.

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

Settled by one-sided unbounded behaviour.

Tracking the sign on the way in

Coming in from the left means $x$ stays negative the whole way: $-0.1$ , then $-0.01$ , then $-0.001$ . A positive numerator over a negative denominator is negative, so every value in this approach sits below 0.

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

The magnitudes match the right-hand side exactly, since $\left|\frac{1}{x}\right| = \frac{1}{|x|}$ depends only on the size of $x$ . Sign is the whole difference.

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

For any bound $M > 0$ , every $x$ with $-\frac{1}{M} < x < 0$ gives $\frac{1}{x} < -M$ . The values drop past every floor you can set, so nothing finite is being approached.

Why substitution fails, and what form actually appears

Substitution gives $\frac{1}{0}$ , undefined, and not the indeterminate $\frac{0}{0}$ . Nothing is competing: the numerator is pinned at 1 while the denominator collapses, which forces unbounded values before any technique gets involved.

That turns the problem into a sign question, and sign questions are answered by the side you came in from. Write down one test input from that side, here a small negative number, and read the sign of the denominator off it.

A two-step habit worth keeping

First fix the magnitude: nonzero over vanishing means unbounded. Then fix the sign with a single test input on the correct side. For $\frac{1}{x-4}$ near 4, testing $3.9$ gives a negative denominator, so $\lim_{x \to 4^-} \frac{1}{x-4} = -\infty$ .

The two sides disagree here, which settles a second question at no extra cost. $\lim_{x \to 0} \frac{1}{x}$ does not exist, because one-sided limits running to opposite infinities do not combine into an infinite two-sided limit.

The mistake students make

Answering $\infty$ because the values are getting big. They are getting large in size and staying negative in sign, and the notation has to carry both facts.
Assuming a reciprocal cannot be negative. $\frac{1}{x}$ carries the same sign as $x$ for every $x \neq 0$ , so the left branch is negative throughout.
Copying the $\frac{1}{x^2}$ answer. The even power there hides the sign of $x$ , so both sides give $\infty$ . The odd power in $\frac{1}{x}$ hides nothing.
Reading the sign off the direction of travel rather than off the input. Coming in from the left does not by itself make an answer negative; a negative denominator does.

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

Does approaching from the left always give a negative answer?

No. The side tells you which inputs to test, not which sign to write. $\lim_{x \to 0^-} \frac{1}{x^2} = \infty$ , because the square is positive on both sides. Test an input from the correct side and read the sign of the denominator.

Can I write DNE instead of negative infinity?

The limit does fail to exist as a real number, but $-\infty$ is the expected answer because it names the behaviour. Free response scoring treats the direction as part of describing the vertical asymptote, so the bare non-existence claim gives up information.

What does this say about the graph?

The line $x = 0$ is a vertical asymptote. The left branch of the hyperbola plunges down along it while the right branch climbs, and the two branches never meet. That picture is the two-sided limit failing, drawn out.