AP Calculus AB and BC

Limit of sec x as x Approaches pi/2 from the Left

Q: Does sec x have a limit at pi/2?

Not two-sided. From the left it is +\infty, from the right -\infty, so the two-sided limit does not exist and x = \pi/2 is a vertical asymptote.

Q: What is the range of secant?

The limit of sec x as x approaches pi over 2 from the left is infinity. Secant is the reciprocal of cosine, and cosine shrinks to zero through positive values on that side, so the reciprocal grows without bound.

\lim_{x \to \frac{\pi}{2}^-} \sec x = \infty

Settled by reciprocal of a vanishing positive quantity.

A reciprocal, nothing more

\sec x = \frac{1}{\cos x}

Secant has no independent behaviour: it is entirely determined by cosine. Wherever cosine vanishes, secant has a vertical asymptote, and wherever cosine is small and positive, secant is large and positive.

Approaching $\pi/2$ from the left keeps the angle in the first quadrant, so $\cos x$ is positive and shrinking. The reciprocal therefore runs to $+\infty$ .

Reading the sign off the quadrant

The fastest reliable check is to name the quadrant. Just left of $\pi/2$ is quadrant I, where cosine is positive. Just right of it is quadrant II, where cosine is negative, so the same reciprocal runs to $-\infty$ .

Secant also never takes a value between $-1$ and 1, which is a useful sanity check: if a secant computation returns $0.4$ , something has gone wrong.

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 sec x have a limit at pi/2?

Not two-sided. From the left it is $+\infty$ , from the right $-\infty$ , so the two-sided limit does not exist and $x = \pi/2$ is a vertical asymptote.

What is the range of secant?