AP Calculus AB and BC

Limit of (sec x - 1) / x at 0 Is 0

Q: What is the limit of (sec x - 1) / x as x approaches 0?

It is 0, because it is the derivative of \sec x at 0, and \sec 0\tan 0 = 0.

Q: Why is this not 1?

The analogy with \frac{e^{x}-1}{x} fails because that limit equals the derivative of e^{x} at 0, which is 1, while the derivative of \sec x at 0 is 0.

Q: What is the limit of (1 - cos x) / x^2?

It is \frac{1}{2}. Changing the power in the denominator changes the answer completely.

The limit of sec x minus 1, over x, as x approaches 0 is 0. This is exactly the definition of the derivative of sec x at 0, and since that derivative is sec x tan x, evaluating at 0 gives 1 times 0, which is 0.

\lim_{x \to 0} \frac{\sec x - 1}{x} = 0

Settled by recognising it as a derivative at a point.

Recognising a derivative in disguise

The shape $\frac{f(x) - f(0)}{x - 0}$ is the alternate form of the derivative at $0$ . With $f(x) = \sec x$ and $f(0) = \sec 0 = 1$ , this limit IS $f'(0)$ .

\lim_{x \to 0}\frac{\sec x - 1}{x} = \left.\frac{d}{dx}\sec x\right|_{x=0} = \sec 0 \tan 0 = 1 \cdot 0 = 0

Spotting the pattern is worth marks

Any limit of the form f(x) minus f(a) over x minus a is a derivative. Recognising it turns a limit problem into a differentiation problem you can do in one line.

The algebraic route

Without the derivative shortcut, rewrite secant as a reciprocal and combine.

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

The first factor is the other standard trigonometric limit and tends to $0$ , while $\frac{1}{\cos x}$ tends to $1$ , so the product is $0$ .

\lim_{x \to 0}\frac{1 - \cos x}{x} = 0

The mistakes students make

Answering $1$ by analogy with $\frac{e^{x}-1}{x}$ . That one is $1$ because the derivative of $e^{x}$ at $0$ is $1$ ; here the derivative of $\sec x$ at $0$ is $0$ .
Dividing $\frac{1-\cos x}{x^{2}}$ and $\frac{1-\cos x}{x}$ up. The first is $\frac{1}{2}$ and the second is $0$ ; the power underneath decides.
Reading $\sec x$ as $\frac{1}{\sin x}$ . Secant is one over COSINE.

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 (sec x - 1) / x as x approaches 0?

It is $0$ , because it is the derivative of $\sec x$ at $0$ , and $\sec 0\tan 0 = 0$ .

Why is this not 1?

The analogy with $\frac{e^{x}-1}{x}$ fails because that limit equals the derivative of $e^{x}$ at $0$ , which is $1$ , while the derivative of $\sec x$ at $0$ is $0$ .

What is the limit of (1 - cos x) / x^2?

It is $\frac{1}{2}$ . Changing the power in the denominator changes the answer completely.