AP Calculus AB and BC

Limit of tan x / sin x as x Approaches 0 Is 1

Q: Do I need the sin x / x limit for this one?

No. The sines cancel outright, and the leftover \frac{1}{\cos x} is continuous at 0. The special limit is for a sine sitting over a plain x, not over another sine.

Q: Is the function continuous at x = 0?

No. Both \tan 0 and \sin 0 are 0, so the expression is undefined there. The discontinuity is removable, and the limit is what tells you the value 1 patches it.

Q: What happens to tan x / sin x near pi/2?

The simplified form \frac{1}{\cos x} has a vertical asymptote there. Coming from the left, \cos x \to 0^+ and the expression tends to +\infty; from the right it tends to -\infty, so the two-sided limit does not exist.

The limit of tan x over sin x as x approaches 0 is 1. Substitution gives 0/0, but here that form is a false alarm: writing tan x as sin x over cos x lets the sines cancel, leaving 1 over cos x, which is continuous at 0 and equals 1 there. No special limit and no L'Hopital required.

\lim_{x \to 0} \frac{\tan x}{\sin x} = 1

Settled by algebraic simplification.

Cancel the sines and substitute

Rewrite the tangent in terms of sine and cosine, then divide by $\sin x$ .

\frac{\tan x}{\sin x} = \frac{\frac{\sin x}{\cos x}}{\sin x} = \frac{\sin x}{\cos x} \cdot \frac{1}{\sin x} = \frac{1}{\cos x}

For every $x \neq 0$ close to 0, $\sin x \neq 0$ , so the cancellation is valid on the whole punctured neighbourhood that a limit examines. What is left is $\sec x$ , which is continuous at 0, so substitution ends the problem.

\lim_{x \to 0} \frac{\tan x}{\sin x} = \lim_{x \to 0} \frac{1}{\cos x} = \frac{1}{1} = 1

A removable discontinuity

The original expression is undefined at $x = 0$ , while $\frac{1}{\cos x}$ is perfectly well behaved there. The two agree at every other point near 0, which is exactly the picture of a removable discontinuity: the graph has a hole at $(0, 1)$ , and defining the value to be 1 patches it.

Why 0/0 does not mean you need a named technique

Substitution gives $\frac{\tan 0}{\sin 0} = \frac{0}{0}$ , so something has to change before the limit can be read off. The something is one line of algebra, not a theorem. Both heavier tools still work: the special limit route rewrites the quotient as $\frac{\sin x}{x} \cdot \frac{x}{\sin x \cos x}$ , and L'Hopital gives $\frac{\sec^2 x}{\cos x} \to 1$ , each with more labour than the problem deserves.

The habit worth building is an order of operations. Substitute first. If the form is indeterminate, simplify: factor, cancel, rationalise, or rewrite everything in sines and cosines. Only when simplification leaves the form intact do you reach for the squeeze theorem or L'Hopital's rule. On this page the second step finishes the job, and CED Topic 1.7 is about making that choice deliberately.

Common mistakes

Treating $\frac{0}{0}$ as an automatic call for L'Hopital. Algebra clears this one in a single step.
Simplifying to $\cos x$ . Dividing by $\sin x$ leaves $\frac{1}{\cos x}$ , not $\cos x$ . The two agree at $x = 0$ , so the limit still comes out right, but every follow-up question about the simplified function goes wrong.
Concluding that the limit fails to exist because the function is undefined at $x = 0$ . A limit never uses the value at the point.
Assuming the answer transfers to relatives. $\frac{\sin x}{\tan x} \to 1$ as well, since it simplifies to $\cos x$ , but $\frac{\tan x}{\sin 2x} \to \frac{1}{2}$ , because it simplifies to $\frac{1}{2\cos^2 x}$ .
Reading the limit at 0 as a statement about the whole domain. $\frac{\tan x}{\sin x}$ is undefined at every multiple of $\pi$ and at every odd multiple of $\frac{\pi}{2}$ , and it is unbounded near the latter.

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

Do I need the sin x / x limit for this one?

No. The sines cancel outright, and the leftover $\frac{1}{\cos x}$ is continuous at 0. The special limit is for a sine sitting over a plain $x$ , not over another sine.

Is the function continuous at x = 0?

No. Both $\tan 0$ and $\sin 0$ are 0, so the expression is undefined there. The discontinuity is removable, and the limit is what tells you the value 1 patches it.

What happens to tan x / sin x near pi/2?

The simplified form $\frac{1}{\cos x}$ has a vertical asymptote there. Coming from the left, $\cos x \to 0^+$ and the expression tends to $+\infty$ ; from the right it tends to $-\infty$ , so the two-sided limit does not exist.