MHT-CET Maths · Limits

Trigonometric Limits — sin x/x and the 1 − cos x Family

Every trigonometric limit at 0 reduces to two facts — sin x/x tends to 1 and (1 − cos x)/x² tends to ½ — once the argument is scaled and the point shifted to 0.

All Limits notes MHT-CET Maths strategy

Why this matters

15 PYQs at 67% HARD — one of the hardest in the chapter, and the page where the difficulty is real rather than clerical. The pattern is fixed: rewrite with an identity, shift the point to 0 if it is not there already, then read off powers of x. Three stems here are third-order — the first-order expansions cancel to 0/0 again — and those are exactly the questions that eat four minutes when you do not know to expand one order further.

Concept 1 of 6: sin x/x, tan x/x and Scaled Arguments

For a tiny angle, the arc and the chord are the same length:

\sin x \approx x

, and equally

\tan x \approx x

. So

\sin(\text{anything small})

may be replaced by that small thing whenever it sits in a ratio.

Definition

$\lim_{x\to 0}\dfrac{\sin x}{x} = 1$ and $\lim_{x\to 0}\dfrac{\tan x}{x} = 1$ ; the reciprocals $\dfrac{x}{\sin x}$ , $\dfrac{x}{\tan x}$ also tend to $1$ . The angle must be in radians and must tend to $0$ .
Scaled argument: $\dfrac{\sin kx}{x} = k\cdot\dfrac{\sin kx}{kx} \to k$ . In general each $\sin(\square)$ or $\tan(\square)$ with $\square \to 0$ may be replaced by $\square$ inside a product or quotient.
$\dfrac{\sin ax}{\sin bx} \to \dfrac{a}{b}$ , $\dfrac{\tan ax}{\sin bx} \to \dfrac{a}{b}$ , $x\cot kx = \dfrac{x\cos kx}{\sin kx} \to \dfrac{1}{k}$ .
Count the powers of $x$ on each floor after replacement. If they match, the limit is the ratio of coefficients; if not, it is $0$ or $\infty$ .
Replacement is only safe in products and quotients. In a difference like $\tan x - \sin x$ the first-order terms cancel, and you need the next order (last concept on this page).

The sine and tangent standard limits

\lim_{x\to 0}\frac{\sin x}{x} = 1 \qquad \lim_{x\to 0}\frac{\tan x}{x} = 1 \qquad \lim_{x\to 0}\frac{\sin kx}{x} = k \qquad \lim_{x\to 0}\frac{\sin ax}{\sin bx} = \frac{a}{b}

Worked example

Evaluate

\lim_{x\to 0}\dfrac{\sin 5x}{\tan 3x}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 12th May Shift 2 · Q127Moderate

Example 1 · Limits · Trigonometric Limits — sin x/x and the 1 − cos x Family

\lim_{x \to 0} \frac{x\cot 4x}{\sin^2 x \cdot \cot^2(2x)}

is equal to

sin x/x → 1 only as x → 0

x \to \infty

\dfrac{\sin x}{x} \to 0

: the numerator stays between

-1

and

1

while the denominator grows. The standard limit is a statement about small angles, not about the function.

Concept 2 of 6: The 1 − cos x Family: (1 − cos kx)/x² = k²/2

1 - \cos x

is second-order small — it behaves like

\dfrac{x^2}{2}

, not like

x

. The identity

1 - \cos x = 2\sin^2\dfrac{x}{2}

is how that half arises: two factors of

\sin(x/2) \approx x/2

Definition

$1 - \cos x = 2\sin^2\dfrac{x}{2}$ , hence $\lim_{x\to 0}\dfrac{1 - \cos x}{x^2} = \dfrac{1}{2}$ .
Scaled: $\lim_{x\to 0}\dfrac{1 - \cos kx}{x^2} = \dfrac{k^2}{2}$ . So $1 - \cos 2x \sim 2x^2$ , $1 - \cos 4x \sim 8x^2$ , $1 - \cos\dfrac{x}{2} \sim \dfrac{x^2}{8}$ .
Consequently $\dfrac{1 - \cos x}{x} \to 0$ and $\dfrac{1 - \cos x}{\sin^2 x} \to \dfrac{1}{2}$ .
$1 - \cos 2x = 2\sin^2 x$ is the same identity with the angle doubled — the form the paper prefers.
A product of two such brackets, $(1 - \cos\frac{x}{2})(1 - \cos\frac{x}{4})$ , is fourth-order: $\dfrac{x^2}{8}\cdot\dfrac{x^2}{32} = \dfrac{x^4}{256}$ .

1 − cos x and its scaling

1 - \cos x = 2\sin^2\frac{x}{2} \qquad \lim_{x\to 0}\frac{1 - \cos x}{x^2} = \frac{1}{2} \qquad \lim_{x\to 0}\frac{1 - \cos kx}{x^2} = \frac{k^2}{2}

Worked example

Evaluate

\lim_{x\to 0}\dfrac{1 - \cos 3x}{x\sin 2x}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 9th May Shift 2 · Q140Moderate

Example 2 · Limits · Trigonometric Limits — sin x/x and the 1 − cos x Family

\lim_{x\to 0}\dfrac{(1-\cos 2x)\cdot\sin 5x}{x^2\sin 3x}

Treating 1 − cos x as first order

\dfrac{1 - \cos x}{x}

0

, not

1

and not

\frac12

. The

\frac12

belongs with

x^2

in the denominator. Count the power of

x

below before you write the constant.

Drill 2 more on the 1 − cos x family: (1 − cos kx)/x² = k²/2

Concept 3 of 6: Rewrite with an Identity Before Taking the Limit

Half the HARD trigonometric limits are hard only because the standard form is hidden behind an identity:

\sin(\pi\cos^2 x)

\sin(\pi\sin^2 x)

in disguise, and

\cos 3x - \cos 5x

is a product of two sines. Do the trigonometry first, the limit second.

Definition

$\sin(\pi - \theta) = \sin\theta$ : so $\sin(\pi\cos^2 x) = \sin(\pi - \pi\sin^2 x) = \sin(\pi\sin^2 x)$ , whose argument does tend to $0$ .
$\cos A - \cos B = -2\sin\dfrac{A + B}{2}\sin\dfrac{A - B}{2}$ : turns a difference of cosines into a product of two small sines, each replaceable by its argument.
$\sqrt{2 - 2\cos\phi} = \sqrt{4\sin^2\frac{\phi}{2}} = 2\left|\sin\dfrac{\phi}{2}\right|$ — note the modulus.
$1 + x\sin x - \cos x = 2\sin^2\dfrac{x}{2} + x\sin x \sim \dfrac{x^2}{2} + x^2 = \dfrac{3x^2}{2}$ : split into pieces you already know the order of.
A factorisable argument such as $x^2 - 12x + 35 = (x - 5)(x - 7)$ is what makes $\dfrac{\sin(\text{that})}{x - 5}$ finite: one factor cancels, the other is evaluated.

Identities that expose a standard form

\sin(\pi - \theta) = \sin\theta \qquad \cos A - \cos B = -2\sin\frac{A + B}{2}\sin\frac{A - B}{2} \qquad \sqrt{2 - 2\cos\phi} = 2\left|\sin\frac{\phi}{2}\right|

Worked example

Evaluate

\lim_{x\to 0}\dfrac{\cos 3x - \cos 5x}{x^2}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 25 April Shift I · Q133Hard

Example 3 · Limits · Trigonometric Limits — sin x/x and the 1 − cos x Family

\lim_{x\rightarrow 5} \frac{\sqrt{2 - 2\cos\left( x^{2}- 12x+ 35 \right)}}{(x- 5)}=

...... .

√(2 − 2cos φ) is 2|sin(φ/2)|, and the modulus decides the sides

\dfrac{\sqrt{2 - 2\cos(x^2 - 12x + 35)}}{x - 5}

the argument

(x - 5)(x - 7)

is negative just right of

5

and positive just left, so the left limit is

-2

and the right limit

+2

. The official key drops the modulus and answers

-2

, as the textbook convention does — on the paper, answer

-2

; in your own understanding, know the two-sided limit does not exist.

cos²x does not tend to 0

\sin(\pi\cos^2 x)

cannot be replaced by

\pi\cos^2 x

— that argument tends to

\pi

, not

0

. Use

\sin(\pi - \theta) = \sin\theta

to move to

\pi\sin^2 x

first.

Drill 1 more on rewrite with an identity before taking the limit

Concept 4 of 6: Shift the Variable: Limits at π/2 and Other Non-Zero Points

The standard limits live at

0

. A limit at

\dfrac{\pi}{2}

is brought home by writing

x = \dfrac{\pi}{2} - h

: as

x \to \dfrac{\pi}{2}

h \to 0

, and every trigonometric function of

x

becomes its co-function of

h

Definition

Put $x = \dfrac{\pi}{2} - h$ . Then $\sin x = \cos h$ , $\cos x = \sin h$ , $\tan x = \cot h$ , $\cot x = \tan h$ , and $\pi - 2x = 2h$ .
Put $x = \dfrac{\pi}{2} + h$ instead if the stem's form suggests it: $\sin x = \cos h$ , $\cos x = -\sin h$ , $\cot x = -\tan h$ , $\pi - 2x = -2h$ . Either works; signs must be tracked.
At $x \to \pi$ : $x = \pi + h$ gives $\sin x = -\sin h$ , $\cos x = -\cos h$ , so $1 + \cos x = 1 - \cos h$ .
At $x \to 1$ with a $\cos(\pi x)$ : $x = 1 - h$ gives $\cos(\pi - \pi h) = -\cos\pi h$ , so $1 + \cos\pi x = 1 - \cos\pi h \sim \dfrac{\pi^2 h^2}{2}$ .
Powers: $(\pi - 2x)^3 = 8h^3$ and $(\pi - 2x)^4 = 16h^4$ — the constant is raised to the power too.
After the shift, everything is a product of $\sin h$ , $1 - \cos h$ , $\tan h$ and powers of $h$ : count orders and read off the constant.

The π/2 shift

x = \tfrac{\pi}{2} - h:\quad \sin x = \cos h,\ \ \cos x = \sin h,\ \ \cot x = \tan h,\ \ \pi - 2x = 2h,\ \ 1 - \sin x = 1 - \cos h \sim \tfrac{h^2}{2}

Worked example

Evaluate

\lim_{x\to\pi/2}\dfrac{1 - \sin x}{(\pi - 2x)^2}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 9th May Shift 1 · Q138Hard

Example 4 · Limits · Trigonometric Limits — sin x/x and the 1 − cos x Family

\lim_{x\to\pi/2}\frac{\cot x-\cos x}{(\pi-2x)^3}

equals

(π − 2x)³ is 8h³, not h³

The factor

2

is cubed along with

h

. Forgetting it multiplies the answer by

8

, and

\dfrac{1}{2}

instead of

\dfrac{1}{16}

is always among the options.

Drill 2 more on shift the variable: limits at π/2 and other non-zero points

Concept 5 of 6: Degrees Are Not Radians

\sin x/x \to 1

is a radian statement. A degree is a much smaller angle —

\dfrac{\pi}{180}

of a radian — so

\sin x^\circ

is roughly

\dfrac{\pi x}{180}

, and every standard limit picks up that factor.

Definition

$x^\circ = \dfrac{\pi x}{180}$ radians. Convert inside every trigonometric function before using any standard limit.
$\lim_{x\to 0}\dfrac{\sin x^\circ}{x} = \dfrac{\pi}{180}$ , $\lim_{x\to 0}\dfrac{\tan x^\circ}{x} = \dfrac{\pi}{180}$ , $\lim_{x\to 0}\dfrac{1 - \cos x^\circ}{x^2} = \dfrac{1}{2}\left(\dfrac{\pi}{180}\right)^2$ .
Differences of cosines in degrees combine both ideas: $\dfrac{\cos mx^\circ - \cos nx^\circ}{x^2} \to \dfrac{(n^2 - m^2)\pi^2}{2\cdot 180^2}$ .
The answer to a degree question always carries $\pi$ and a power of $180$ (or a divisor of it); an answer without $\pi$ has ignored the degree sign.

Degree conversion in a limit

x^\circ = \frac{\pi x}{180}\ \text{rad} \qquad \lim_{x\to 0}\frac{\sin x^\circ}{x} = \frac{\pi}{180} \qquad \lim_{x\to 0}\frac{1 - \cos x^\circ}{x^2} = \frac{\pi^2}{2\cdot 180^2}

Worked example

Evaluate

\lim_{x\to 0}\dfrac{\sin x^\circ}{x}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 12th May Shift 1 · Q144Hard

Example 5 · Limits · Trigonometric Limits — sin x/x and the 1 − cos x Family

\displaystyle\lim_{x\to0}\frac{\cos7x^{\circ}-\cos2x^{\circ}}{x^{2}}

Dropping the degree sign

\dfrac{\cos 7x^\circ - \cos 2x^\circ}{x^2}

-\dfrac{\pi^2}{1440}

; the radian version would be

-\dfrac{45}{2}

, which is the first distractor. If the stem prints a small circle, the answer has a

\pi

in it.

Concept 6 of 6: Higher-Order Forms: Expand to the Needed Power

When the first-order pieces cancel —

\tan x - \sin x

x\tan 2x - 2x\tan x

— the limit lives in the

x^3

terms. Carrying one more term of each series is faster and safer than three rounds of L'Hôpital.

Definition

Series to memorise: $\sin x = x - \dfrac{x^3}{6} + \dots$ , $\tan x = x + \dfrac{x^3}{3} + \dots$ , $\cos x = 1 - \dfrac{x^2}{2} + \dfrac{x^4}{24} - \dots$ , $e^x = 1 + x + \dfrac{x^2}{2} + \dots$ , $\log(1 + x) = x - \dfrac{x^2}{2} + \dots$ .
Rule: expand every term to the order of the denominator, and no further. A denominator of $x^3$ needs cubic terms; $x^4$ needs quartic terms.
Scaled arguments scale the series: $\tan 2x = 2x + \dfrac{(2x)^3}{3} = 2x + \dfrac{8x^3}{3}$ .
Differences of standard forms: $\tan x - \sin x \sim \dfrac{x^3}{2}$ , $x - \sin x \sim \dfrac{x^3}{6}$ , $\tan x - x \sim \dfrac{x^3}{3}$ , $e^x - 1 - x \sim \dfrac{x^2}{2}$ .
Without series: $\tan h - \sin h = \tan h\,(1 - \cos h)$ — a product of first- and second-order pieces, which is often the cleaner route on the paper.

Series to the third order

\sin x = x - \frac{x^3}{6} + \dots \qquad \tan x = x + \frac{x^3}{3} + \dots \qquad \cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \dots \qquad e^x = 1 + x + \frac{x^2}{2} + \dots

Worked example

Evaluate

\lim_{x\to 0}\dfrac{\tan x - \sin x}{x^3}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 11th May Shift 2 · Q111Hard

Example 6 · Limits · Trigonometric Limits — sin x/x and the 1 − cos x Family

\lim_{x\to 0}\frac{x\tan 2x-2x\tan x}{(1-\cos 2x)^2}

Stopping at first order and getting 0/0 again

\dfrac{x\tan 2x - 2x\tan x}{(1 - \cos 2x)^2}

: first order gives

2x^2 - 2x^2 = 0

on top. Go to third order —

x\left(2x + \frac{8x^3}{3}\right) - 2x\left(x + \frac{x^3}{3}\right) = 2x^4

— against

(2x^2)^2 = 4x^4

below, and the limit is

\dfrac{1}{2}

. The cancellation is the signal to expand further, not to answer

0

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (6)

sin x/x, tan x/x and Scaled Arguments
The sine and tangent standard limits
$\lim_{x\to 0}\frac{\sin x}{x} = 1 \qquad \lim_{x\to 0}\frac{\tan x}{x} = 1 \qquad \lim_{x\to 0}\frac{\sin kx}{x} = k \qquad \lim_{x\to 0}\frac{\sin ax}{\sin bx} = \frac{a}{b}$
The 1 − cos x Family: (1 − cos kx)/x² = k²/2
1 − cos x and its scaling
$1 - \cos x = 2\sin^2\frac{x}{2} \qquad \lim_{x\to 0}\frac{1 - \cos x}{x^2} = \frac{1}{2} \qquad \lim_{x\to 0}\frac{1 - \cos kx}{x^2} = \frac{k^2}{2}$
Rewrite with an Identity Before Taking the Limit
Identities that expose a standard form
$\sin(\pi - \theta) = \sin\theta \qquad \cos A - \cos B = -2\sin\frac{A + B}{2}\sin\frac{A - B}{2} \qquad \sqrt{2 - 2\cos\phi} = 2\left|\sin\frac{\phi}{2}\right|$
Shift the Variable: Limits at π/2 and Other Non-Zero Points
The π/2 shift
$x = \tfrac{\pi}{2} - h:\quad \sin x = \cos h,\ \ \cos x = \sin h,\ \ \cot x = \tan h,\ \ \pi - 2x = 2h,\ \ 1 - \sin x = 1 - \cos h \sim \tfrac{h^2}{2}$
Degrees Are Not Radians
Degree conversion in a limit
$x^\circ = \frac{\pi x}{180}\ \text{rad} \qquad \lim_{x\to 0}\frac{\sin x^\circ}{x} = \frac{\pi}{180} \qquad \lim_{x\to 0}\frac{1 - \cos x^\circ}{x^2} = \frac{\pi^2}{2\cdot 180^2}$
Higher-Order Forms: Expand to the Needed Power
Series to the third order
$\sin x = x - \frac{x^3}{6} + \dots \qquad \tan x = x + \frac{x^3}{3} + \dots \qquad \cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \dots \qquad e^x = 1 + x + \frac{x^2}{2} + \dots$

Watch out for (7)

sin x/x → 1 only as x → 0
→ sin x/x, tan x/x and Scaled Arguments
Treating 1 − cos x as first order
→ The 1 − cos x Family: (1 − cos kx)/x² = k²/2
√(2 − 2cos φ) is 2|sin(φ/2)|, and the modulus decides the sides
→ Rewrite with an Identity Before Taking the Limit
cos²x does not tend to 0
→ Rewrite with an Identity Before Taking the Limit
(π − 2x)³ is 8h³, not h³
→ Shift the Variable: Limits at π/2 and Other Non-Zero Points
Dropping the degree sign
→ Degrees Are Not Radians
Stopping at first order and getting 0/0 again
→ Higher-Order Forms: Expand to the Needed Power

Test yourself on Limits

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