AP Calculus AB and BC

Limit of csc x as x Approaches 0 from the Right

Q: Why is this not an indeterminate form?

Indeterminate forms need BOTH parts to misbehave, as in 0/0 or \infty/\infty. Here the numerator is a fixed 1, so the answer is determined: unbounded, with the sign set by the denominator.

Q: Where else does cosecant have asymptotes?

At every integer multiple of \pi, since those are the zeros of sine.

The limit of csc x as x approaches 0 from the right is infinity. Cosecant is one over sine, and just to the right of zero sine is small and positive, so its reciprocal grows without bound.

\lim_{x \to 0^+} \csc x = \infty

Settled by reciprocal of a vanishing positive quantity.

The sign is the whole problem

\csc x = \frac{1}{\sin x}

For small positive $x$ , $\sin x \approx x > 0$ , so $\csc x \approx 1/x$ , which runs to $+\infty$ . For small negative $x$ , $\sin x \approx x < 0$ and the reciprocal runs to $-\infty$ .

So the two-sided limit does not exist, and the origin is a vertical asymptote of the cosecant graph.

Do not confuse it with the special limit

The famous limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$ can make cosecant near 0 look tame. It is not. That limit is a quotient of two quantities BOTH tending to 0, an indeterminate form that happens to resolve to 1.

Here there is no indeterminate form at all: the numerator is the constant 1 and only the denominator vanishes. A nonzero constant over something vanishing is never indeterminate, it is unbounded.

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 this not an indeterminate form?

Indeterminate forms need BOTH parts to misbehave, as in $0/0$ or $\infty/\infty$ . Here the numerator is a fixed 1, so the answer is determined: unbounded, with the sign set by the denominator.

Where else does cosecant have asymptotes?

At every integer multiple of $\pi$ , since those are the zeros of sine.