AP Calculus AB and BC

Limit of x/(x-1) as x Approaches 1 from the Right

Q: Is 1 over 0 an indeterminate form?

No. Only 0/0, \infty/\infty, 0 \cdot \infty, \infty - \infty, 0^{0}, 1^{\infty} and \infty^{0} are indeterminate. A nonzero over zero is decided: unbounded.

The limit of x over x minus one as x approaches one from the right is infinity. The numerator tends to one, a fixed nonzero value, while the denominator shrinks to zero through positive values, so the quotient grows without bound.

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

Settled by one-sided analysis at a vertical asymptote.

Check the numerator first

Substituting $x = 1$ gives $1/0$ , which is NOT indeterminate. A nonzero over a zero is always unbounded; the only work left is finding the sign.

Compare with $\frac{x-1}{x-1}$ , where substitution gives $0/0$ and the answer is 1. Reading which of the two you have, before doing anything else, is what stops a wasted L'Hopital attempt.

Getting the sign

Approaching from the right means $x > 1$ , so $x - 1$ is a small POSITIVE number. The numerator is close to 1, also positive. Positive over small positive is large and positive.

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

The sides disagree, so the two-sided limit does not exist and $x = 1$ is a vertical asymptote.

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 decide the sign of an infinite limit?

Pick a test value very close to the approach point on the correct side and check the sign of the numerator and denominator separately. $x = 1.001$ gives roughly $1/0.001$ , positive.

Is 1 over 0 an indeterminate form?

No. Only $0/0$ , $\infty/\infty$ , $0 \cdot \infty$ , $\infty - \infty$ , $0^{0}$ , $1^{\infty}$ and $\infty^{0}$ are indeterminate. A nonzero over zero is decided: unbounded.