AP Calculus AB and BC

Limit of 1/(x - 1) at 1 From the Right Is Infinity

Q: What is the limit of 1/(x-1) as x approaches 1 from the right?

It is \infty. The denominator x-1 is positive and shrinking, so the quotient grows past every bound.

Q: Why is the left-hand limit different?

Below 1 the denominator x-1 is negative, so the quotient is a large negative number and \lim_{x \to 1^{-}}\frac{1}{x-1} = -\infty.

Q: Does the two-sided limit exist?

No. The two one-sided limits are \infty and -\infty, so \lim_{x \to 1}\frac{1}{x-1} does not exist.

The limit of 1 over x minus 1 as x approaches 1 from the right is infinity. Just above 1 the denominator is a small positive number, so the quotient grows without bound. Just below 1 it is small and negative, so the two-sided limit does not exist.

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

Settled by one-sided analysis at a vertical asymptote.

Approaching from the right keeps the denominator positive

Substitution gives $\frac{1}{0}$ , which is not a value but a signal: the size of the quotient runs away. All that is left to decide is the sign, and the side you come in from decides it.

For $x$ just above $1$ , the difference $x-1$ is a small positive number, so the quotient is a large positive number.

\frac{1}{1.01-1} = 100, \qquad \frac{1}{1.0001-1} = 10000

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

The left side, and the two-sided verdict

Coming in from below, $x = 0.999$ makes $x-1$ equal to $-0.001$ , and the quotient is $-1000$ . Values keep the same sign and keep growing in size, so the left-hand limit runs to negative infinity.

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

The two sides disagree, so $\lim_{x \to 1}\frac{1}{x-1}$ does not exist. The line $x = 1$ is a vertical asymptote, and writing $\infty$ records unbounded growth rather than a number the function reaches.

Read the superscript first

On a problem like this the algebra is over before it starts. The entire question is which side of 1 the values come from, so find the small plus or minus first, then test one number on that side.

The mistakes students make

Nearly every error here is a sign error or a superscript that went unread.

Answering $-\infty$ after testing $x = 0.999$ . That value sits to the left of $1$ , and the superscript plus asks for values above $1$ .
Answering that the limit does not exist. The two-sided limit does not exist, but the right-hand limit is $\infty$ , and the right-hand limit is what was asked for.
Reading $\frac{1}{0}$ as $0$ . Dividing by a quantity heading to zero makes the result enormous, not small.

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

What is the limit of 1/(x-1) as x approaches 1 from the right?

It is $\infty$ . The denominator $x-1$ is positive and shrinking, so the quotient grows past every bound.

Why is the left-hand limit different?

Below $1$ the denominator $x-1$ is negative, so the quotient is a large negative number and $\lim_{x \to 1^{-}}\frac{1}{x-1} = -\infty$ .

Does the two-sided limit exist?

No. The two one-sided limits are $\infty$ and $-\infty$ , so $\lim_{x \to 1}\frac{1}{x-1}$ does not exist.