Limits of trigonometric functions

Trigonometric functions oscillate, so their limits need one extra tool beyond plain substitution. Near almost any point $a$ , $\sin x$ and $\cos x$ behave predictably, and $\lim_{x\to a}\sin x=\sin a$ , $\lim_{x\to a}\cos x=\cos a$ — direct substitution just works. The interesting case is when the angle itself shrinks to zero and substitution collapses to $\frac00$ . That is where the single most important result in this chapter enters: $\lim_{x\to 0}\frac{\sin x}{x}=1$ , with $x$ measured in radians. This article builds that result from the unit circle, shows how every companion trigonometric limit — $\frac{\tan x}{x}$ , $\frac{1-\cos x}{x}$ , $\frac{\sin kx}{x}$ — falls out of it by algebraic rewriting, and works through the two standard $\frac00$ patterns CBSE Class 11 expects: ratios of sines and expressions built from $1-\cos\theta$ .

The core insight

Every trigonometric $\frac00$ limit near $x=0$ reduces, after rewriting with identities, to the single geometric fact $\lim_{x\to 0}\frac{\sin x}{x}=1$ .

The core idea

Where direct substitution already works. For most points $a$ , trigonometric functions are defined and continuous, so

\lim_{x\to a}\sin x = \sin a, \qquad \lim_{x\to a}\cos x = \cos a,

where: $a$ is any real number, and the same holds for $\tan x$ provided $\cos a \neq 0$ (so the function is actually defined there). If substitution gives a finite answer, stop — no identity manipulation is needed.

Where it breaks down. Substitution fails only when it produces $\frac00$ , almost always because the angle multiplying $x$ shrinks to $0$ together with $x$ . The building-block fact, proved geometrically by comparing the areas of two triangles and a circular sector inside a unit circle, is

\lim_{x\to 0}\frac{\sin x}{x} = 1 \qquad (x \text{ in radians}).

This squeeze — the sector area sits between the two triangle areas — pins $\frac{\sin x}{x}$ between $\cos x$ and $1$ near $0$ , and since both bounds tend to $1$ , so must $\frac{\sin x}{x}$ .

Companion limits fall out for free. Once the base result is fixed, algebra of limits gives the rest without new geometry:

\lim_{x\to 0}\frac{\tan x}{x} = 1, \qquad \lim_{x\to 0}\frac{1-\cos x}{x} = 0, \qquad \lim_{x\to 0}\frac{\sin kx}{x} = k,

where: $k$ is any nonzero constant. For instance $\frac{\tan x}{x}=\frac{\sin x}{x}\cdot\frac{1}{\cos x}\to 1\cdot 1=1$ since $\cos x \to \cos 0 = 1$ .

The general strategy. First try direct substitution. If it gives $\frac00$ , use a compound-angle or sum-to-product identity to expose a factor shaped like $\frac{\sin(\text{something}\to 0)}{\text{something}}$ , then apply the standard limit to each such factor. Three signals that a limit belongs to this family: the expression contains $\sin$ , $\cos$ or $\tan$ of something tending to $0$ ; direct substitution produces $\frac00$ rather than a finite value; and after simplifying, every leftover piece is either a constant or a ratio of the shape $\frac{\sin(\cdot)}{(\cdot)}$ with matching arguments top and bottom. Once all three hold, the answer is just the product of the constants that survive, since every genuine $\frac{\sin(\cdot)}{(\cdot)}$ factor contributes exactly $1$ .

Figure

A unit circle showing a small angle x with the sector area trapped between two triangle areas, illustrating why sin x over x tends to 1

Derivation AWhy sin x over x tends to 1 as x tends to 0

The goal is to show, using only areas of simple shapes drawn inside a unit circle and the squeeze (sandwich) theorem, that $\lim_{x\to 0}\dfrac{\sin x}{x}=1$ when $x$ is a small positive angle measured in radians — the single fact every trigonometric $\frac00$ limit in this chapter is eventually reduced to.

Step 1. Draw a unit circle centred at the origin and a small angle $x$ (in radians, $0 < x < \pi/2$ ) at the centre. Mark three regions sharing the same angle $x$ : the triangle formed by the two radii and the chord (area $\frac12\sin x$ , since two sides of length $1$ enclose angle $x$ ), the circular sector swept by the angle (area $\frac12 x$ , since a full circle of radius $1$ has area $\pi$ and the sector is the fraction $x/2\pi$ of it), and the larger triangle formed by the radius, the tangent line, and the extended radius (area $\frac12\tan x$ ).

Step 2. Compare the three areas geometrically. The small triangle sits entirely inside the sector, and the sector sits entirely inside the large triangle, so their areas satisfy $\frac12\sin x \le \frac12 x \le \frac12\tan x$ .

Step 3. Multiply through by $2$ and divide by $\sin x$ (positive for small $x>0$ ), which reverses no inequality since $\sin x>0$ : $\sin x \le x \le \tan x$ becomes $1 \le \dfrac{x}{\sin x} \le \dfrac{1}{\cos x}$ , and taking reciprocals (which flips the inequalities) gives $\cos x \le \dfrac{\sin x}{x} \le 1$ .

Step 4. Apply the squeeze theorem. As $x \to 0^+$ , $\cos x \to \cos 0 = 1$ , and the constant upper bound is already $1$ , so the trapped quantity $\dfrac{\sin x}{x}$ is squeezed to $1$ . The same bound holds for $x\to 0^-$ by symmetry, since $\frac{\sin x}{x}$ is an even function of $x$ . Hence $\lim_{x\to 0}\dfrac{\sin x}{x}=1$ .

Worked example

Evaluate $\displaystyle\lim_{x \to 0} \frac{\sin 5x}{\sin 3x}$ .

Step 1. Check direct substitution: at $x=0$ the numerator is $\sin 0=0$ and the denominator is $\sin 0=0$ , giving $\frac00$ , so the standard limit must be engineered in.

Step 2. Multiply and divide to expose the standard pattern for each sine separately:

\frac{\sin 5x}{\sin 3x} = \frac{\sin 5x}{5x}\cdot\frac{5x}{3x}\cdot\frac{3x}{\sin 3x} = \frac{\sin 5x}{5x}\cdot\frac{5}{3}\cdot\frac{3x}{\sin 3x}.

Step 3. As $x\to 0$ , both $5x\to 0$ and $3x\to 0$ , so each sine-over-angle factor tends to $1$ by the derivation above: $\lim_{x\to0}\frac{\sin 5x}{5x}=1$ and $\lim_{x\to0}\frac{3x}{\sin 3x}=1$ .

Step 4. Combine with the product rule of limits: $\displaystyle\lim_{x\to 0}\frac{\sin 5x}{\sin 3x} = 1\cdot\frac{5}{3}\cdot 1 = \frac{5}{3}$ . The limit equals $\dfrac{5}{3}$ .

Competitive stretch

Evaluate $\displaystyle\lim_{x \to 0} \frac{1 - \cos 2x}{x^2}$ . Substitution gives $\frac{1-\cos 0}{0}=\frac00$ , so rewrite the numerator with $1-\cos\theta = 2\sin^2(\theta/2)$ at $\theta=2x$ : $1-\cos 2x = 2\sin^2 x$ . Then $\displaystyle\frac{1-\cos 2x}{x^2} = 2\left(\frac{\sin x}{x}\right)^2 \to 2(1)^2 = 2$ . The limit equals $2$ . Notice the general pattern: any limit built from $1-\cos(mx)$ over a power of $x$ collapses the same way — swap in the half-angle identity, split off a squared sine-over-angle factor, and let the standard limit finish the job.

Common traps

Applying $\lim_{x\to0}\frac{\sin x}{x}=1$ to $\frac{\sin kx}{x}$ without first matching the angle in numerator and denominator — they must be the identical expression before the ratio equals $1$ .
Forgetting the standard limit needs radians; using degree measure gives a wrong numerical constant.
Assuming every trigonometric limit is $\frac00$ and needs an identity, when the function is often already defined at the point, so direct substitution suffices.
Misplacing the $\tfrac12$ inside $1-\cos\theta = 2\sin^2(\theta/2)$ , which corrupts the final constant.
Splitting $\frac{\sin ax}{\sin bx}$ as $\frac{\sin ax}{ax}\div\frac{\sin bx}{bx}$ and then losing track of the leftover factor $\frac{a}{b}$ from cancelling the $x$ 's — always write out the multiply-and-divide step explicitly rather than doing it mentally.

Ask the tutor: If a trigonometric limit will not simplify into the sin-over-angle pattern, ask the AI tutor to walk through the multiply-and-divide trick on a fresh example.

Why does the standard limit require radians specifically?

The geometric proof compares the sector area $\frac12 x$ with triangle areas, and that formula for sector area only equals $\frac12 x$ when $x$ is in radians (since radian measure is defined so that arc length equals radius times angle). In degrees the sector area picks up an extra conversion factor, so $\lim_{x\to0}\frac{\sin x}{x}$ would come out to $\frac{\pi}{180}$ , not $1$ . CBSE always assumes radians in calculus unless stated otherwise.

How do I know when to use the standard limit versus just substituting?

Substitute first, always. If the function is defined and continuous at the point — meaning no denominator vanishes and no forbidden value like $\cos(\pi/2)$ for $\tan x$ is hit — the substituted value is the answer, and you never need the standard limit. Reach for $\lim_{x\to0}\frac{\sin x}{x}=1$ only after substitution genuinely produces $\frac00$ .

Is lim tan x over x equal to 1 a separate fact I need to memorise?

No — treat it as a two-line derivation from the sine limit rather than a fact to memorise. Write $\frac{\tan x}{x}=\frac{\sin x}{x}\cdot\frac{1}{\cos x}$ ; as $x\to0$ , the first factor tends to $1$ (standard limit) and the second tends to $\frac{1}{\cos 0}=1$ , so the product tends to $1$ . This rewrite-then-apply pattern handles almost every companion limit in this chapter — the same trick with $\cot x$ , $\sec x$ or $\csc x$ always comes down to rewriting in terms of $\sin x$ and $\cos x$ first and then splitting into pieces you already know how to take the limit of. The moment you see a trigonometric ratio you don't immediately recognise, converting it to sines and cosines is almost always the fastest way forward, since $\lim_{x\to0}\cos x = 1$ never causes trouble on its own.

Limits of trigonometric functions

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads