AP Calculus AB and BC

Limit of tan(x)/(x-pi) as x Approaches pi

Q: Why is the period of tangent pi rather than 2pi?

Because sine and cosine both flip sign over a shift of \pi, and the two sign flips cancel in their quotient. The tangent graph repeats twice as often as the sine graph.

Q: Does the same trick work at other multiples of pi?

Yes. At any integer multiple of \pi the tangent is zero and the same shift applies, so the limit is 1 at every one of them.

The limit of tangent x over x minus pi as x approaches pi is one. Substituting u for x minus pi and using the fact that tangent repeats every pi turns the whole thing into the standard limit of tangent u over u.

\lim_{x \to \pi} \frac{\tan x}{x-\pi} = 1

Settled by shifting with the period, then the standard limit.

Shift the variable

Substitution gives $\frac{\tan \pi}{0} = \frac{0}{0}$ , an indeterminate form. Set $u = x - \pi$ , so $u \to 0$ , and use the period of the tangent.

\tan(u+\pi) = \tan u \quad\Longrightarrow\quad \frac{\tan x}{x-\pi} = \frac{\tan u}{u} \longrightarrow 1

Tangent has period $\pi$ , not $2\pi$ . That is what makes the shift exact rather than merely approximate, and it is the fact this problem is really testing.

Reading it as a derivative

The expression is also the difference quotient for $f(x) = \tan x$ at $x = \pi$ , since $\tan \pi = 0$ . So the limit is $f'(\pi) = \sec^{2}\pi = (-1)^{2} = 1$ .

Both routes give 1. The derivative route is faster if you already know $\frac{d}{dx}\tan x = \sec^{2}x$ ; the substitution route is the one that works before the derivative has been established.

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 the period of tangent pi rather than 2pi?

Because sine and cosine both flip sign over a shift of $\pi$ , and the two sign flips cancel in their quotient. The tangent graph repeats twice as often as the sine graph.

Does the same trick work at other multiples of pi?

Yes. At any integer multiple of $\pi$ the tangent is zero and the same shift applies, so the limit is 1 at every one of them.