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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 notesMHT-CET Maths strategy

Why this matters

12 PYQs at 67% HARD — the hardest page in the chapter, and the one 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

Intuition

For a tiny angle, the arc and the chord are the same length: sin⁡x≈x\sin x \approx x, and equally tan⁡x≈x\tan x \approx x. So sin⁡(anything small)\sin(\text{anything small}) may be replaced by that small thing whenever it sits in a ratio.

Definition

  • lim⁡x→0sin⁡xx=1\lim_{x\to 0}\dfrac{\sin x}{x} = 1 and lim⁡x→0tan⁡xx=1\lim_{x\to 0}\dfrac{\tan x}{x} = 1; the reciprocals xsin⁡x\dfrac{x}{\sin x}, xtan⁡x\dfrac{x}{\tan x} also tend to 11. The angle must be in radians and must tend to 00.
  • Scaled argument: sin⁡kxx=k⋅sin⁡kxkx→k\dfrac{\sin kx}{x} = k\cdot\dfrac{\sin kx}{kx} \to k. In general each sin⁡(□)\sin(\square) or tan⁡(□)\tan(\square) with □→0\square \to 0 may be replaced by □\square inside a product or quotient.
  • sin⁡axsin⁡bx→ab\dfrac{\sin ax}{\sin bx} \to \dfrac{a}{b}, tan⁡axsin⁡bx→ab\dfrac{\tan ax}{\sin bx} \to \dfrac{a}{b}, xcot⁡kx=xcos⁡kxsin⁡kx→1kx\cot kx = \dfrac{x\cos kx}{\sin kx} \to \dfrac{1}{k}.
  • Count the powers of xx on each floor after replacement. If they match, the limit is the ratio of coefficients; if not, it is 00 or ∞\infty.
  • Replacement is only safe in products and quotients. In a difference like tan⁡x−sin⁡x\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→0sin⁡xx=1lim⁡x→0tan⁡xx=1lim⁡x→0sin⁡kxx=klim⁡x→0sin⁡axsin⁡bx=ab\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→0sin⁡5xtan⁡3x\lim_{x\to 0}\dfrac{\sin 5x}{\tan 3x}.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1LimitsHARD
lim⁡x→0xcot⁡4xsin⁡2x⋅cot⁡2(2x)\lim_{x\to 0} \frac{x\cot 4x}{\sin^2 x\cdot\cot^2(2x)} is equal to

[Q127 · 13th May Shift 1 · 2024]

sin x/x → 1 only as x → 0

As x→∞x \to \infty, sin⁡xx→0\dfrac{\sin x}{x} \to 0: the numerator stays between −1-1 and 11 while the denominator grows. The standard limit is a statement about small angles, not about the function.
Drill 1 more on sin x/x, tan x/x and scaled arguments

Concept 2 of 6

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

Intuition

1−cos⁡x1 - \cos x is second-order small — it behaves like x22\dfrac{x^2}{2}, not like xx. The identity 1−cos⁡x=2sin⁡2x21 - \cos x = 2\sin^2\dfrac{x}{2} is how that half arises: two factors of sin⁡(x/2)≈x/2\sin(x/2) \approx x/2.

Definition

  • 1−cos⁡x=2sin⁡2x21 - \cos x = 2\sin^2\dfrac{x}{2}, hence lim⁡x→01−cos⁡xx2=12\lim_{x\to 0}\dfrac{1 - \cos x}{x^2} = \dfrac{1}{2}.
  • Scaled: lim⁡x→01−cos⁡kxx2=k22\lim_{x\to 0}\dfrac{1 - \cos kx}{x^2} = \dfrac{k^2}{2}. So 1−cos⁡2x∼2x21 - \cos 2x \sim 2x^2, 1−cos⁡4x∼8x21 - \cos 4x \sim 8x^2, 1−cos⁡x2∼x281 - \cos\dfrac{x}{2} \sim \dfrac{x^2}{8}.
  • Consequently 1−cos⁡xx→0\dfrac{1 - \cos x}{x} \to 0 and 1−cos⁡xsin⁡2x→12\dfrac{1 - \cos x}{\sin^2 x} \to \dfrac{1}{2}.
  • 1−cos⁡2x=2sin⁡2x1 - \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⁡x2)(1−cos⁡x4)(1 - \cos\frac{x}{2})(1 - \cos\frac{x}{4}), is fourth-order: x28⋅x232=x4256\dfrac{x^2}{8}\cdot\dfrac{x^2}{32} = \dfrac{x^4}{256}.

1 − cos x and its scaling

1−cos⁡x=2sin⁡2x2lim⁡x→01−cos⁡xx2=12lim⁡x→01−cos⁡kxx2=k221 - \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→01−cos⁡3xxsin⁡2x\lim_{x\to 0}\dfrac{1 - \cos 3x}{x\sin 2x}.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2LimitsMODERATE
lim⁡x→0(1−cos⁡2x)⋅sin⁡5xx2sin⁡3x\lim_{x\to 0}\dfrac{(1-\cos 2x)\cdot\sin 5x}{x^2\sin 3x} is

[Q140 · 9th May Shift 2 · 2024]

Treating 1 − cos x as first order

1−cos⁡xx\dfrac{1 - \cos x}{x} is 00, not 11 and not 12\frac12. The 12\frac12 belongs with x2x^2 in the denominator. Count the power of xx 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

Intuition

Half the HARD trigonometric limits are hard only because the standard form is hidden behind an identity: sin⁡(πcos⁡2x)\sin(\pi\cos^2 x) is sin⁡(πsin⁡2x)\sin(\pi\sin^2 x) in disguise, and cos⁡3x−cos⁡5x\cos 3x - \cos 5x is a product of two sines. Do the trigonometry first, the limit second.

Definition

  • sin⁡(π−θ)=sin⁡θ\sin(\pi - \theta) = \sin\theta: so sin⁡(πcos⁡2x)=sin⁡(π−πsin⁡2x)=sin⁡(πsin⁡2x)\sin(\pi\cos^2 x) = \sin(\pi - \pi\sin^2 x) = \sin(\pi\sin^2 x), whose argument does tend to 00.
  • cos⁡A−cos⁡B=−2sin⁡A+B2sin⁡A−B2\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.
  • 2−2cos⁡ϕ=4sin⁡2ϕ2=2∣sin⁡ϕ2∣\sqrt{2 - 2\cos\phi} = \sqrt{4\sin^2\frac{\phi}{2}} = 2\left|\sin\dfrac{\phi}{2}\right| — note the modulus.
  • 1+xsin⁡x−cos⁡x=2sin⁡2x2+xsin⁡x∼x22+x2=3x221 + 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 x2−12x+35=(x−5)(x−7)x^2 - 12x + 35 = (x - 5)(x - 7) is what makes sin⁡(that)x−5\dfrac{\sin(\text{that})}{x - 5} finite: one factor cancels, the other is evaluated.

Identities that expose a standard form

sin⁡(π−θ)=sin⁡θcos⁡A−cos⁡B=−2sin⁡A+B2sin⁡A−B22−2cos⁡ϕ=2∣sin⁡ϕ2∣\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→0cos⁡3x−cos⁡5xx2\lim_{x\to 0}\dfrac{\cos 3x - \cos 5x}{x^2}.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3LimitsHARD
lim⁡x→52−2cos⁡(x2−12x+35)(x−5)=\lim_{x\rightarrow 5} \frac{\sqrt{2 - 2\cos\left( x^{2}- 12x+ 35 \right)}}{(x- 5)}=...... .

[Q133 · 25 April Shift I · 2025]

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

In 2−2cos⁡(x2−12x+35)x−5\dfrac{\sqrt{2 - 2\cos(x^2 - 12x + 35)}}{x - 5} the argument (x−5)(x−7)(x - 5)(x - 7) is negative just right of 55 and positive just left, so the left limit is −2-2 and the right limit +2+2. The official key drops the modulus and answers −2-2, as the textbook convention does — on the paper, answer −2-2; in your own understanding, know the two-sided limit does not exist.

cos²x does not tend to 0

sin⁡(πcos⁡2x)\sin(\pi\cos^2 x) cannot be replaced by πcos⁡2x\pi\cos^2 x — that argument tends to π\pi, not 00. Use sin⁡(π−θ)=sin⁡θ\sin(\pi - \theta) = \sin\theta to move to πsin⁡2x\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

Intuition

The standard limits live at 00. A limit at π2\dfrac{\pi}{2} is brought home by writing x=π2−hx = \dfrac{\pi}{2} - h: as x→π2x \to \dfrac{\pi}{2}, h→0h \to 0, and every trigonometric function of xx becomes its co-function of hh.

Definition

  • Put x=π2−hx = \dfrac{\pi}{2} - h. Then sin⁡x=cos⁡h\sin x = \cos h, cos⁡x=sin⁡h\cos x = \sin h, tan⁡x=cot⁡h\tan x = \cot h, cot⁡x=tan⁡h\cot x = \tan h, and π−2x=2h\pi - 2x = 2h.
  • Put x=π2+hx = \dfrac{\pi}{2} + h instead if the stem's form suggests it: sin⁡x=cos⁡h\sin x = \cos h, cos⁡x=−sin⁡h\cos x = -\sin h, cot⁡x=−tan⁡h\cot x = -\tan h, π−2x=−2h\pi - 2x = -2h. Either works; signs must be tracked.
  • At x→πx \to \pi: x=π+hx = \pi + h gives sin⁡x=−sin⁡h\sin x = -\sin h, cos⁡x=−cos⁡h\cos x = -\cos h, so 1+cos⁡x=1−cos⁡h1 + \cos x = 1 - \cos h.
  • At x→1x \to 1 with a cos⁡(πx)\cos(\pi x): x=1−hx = 1 - h gives cos⁡(π−πh)=−cos⁡πh\cos(\pi - \pi h) = -\cos\pi h, so 1+cos⁡πx=1−cos⁡πh∼π2h221 + \cos\pi x = 1 - \cos\pi h \sim \dfrac{\pi^2 h^2}{2}.
  • Powers: (π−2x)3=8h3(\pi - 2x)^3 = 8h^3 and (π−2x)4=16h4(\pi - 2x)^4 = 16h^4 — the constant is raised to the power too.
  • After the shift, everything is a product of sin⁡h\sin h, 1−cos⁡h1 - \cos h, tan⁡h\tan h and powers of hh: count orders and read off the constant.

The π/2 shift

x=π2−h:sin⁡x=cos⁡h,  cos⁡x=sin⁡h,  cot⁡x=tan⁡h,  π−2x=2h,  1−sin⁡x=1−cos⁡h∼h22x = \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→π/21−sin⁡x(π−2x)2\lim_{x\to\pi/2}\dfrac{1 - \sin x}{(\pi - 2x)^2}.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 4LimitsHARD
lim⁡x→π/2cot⁡x−cos⁡x(π−2x)3\lim_{x\to\pi/2}\frac{\cot x-\cos x}{(\pi-2x)^3} equals

[Q138 · 9th May Shift 1 · 2024]

(π − 2x)³ is 8h³, not h³

The factor 22 is cubed along with hh. Forgetting it multiplies the answer by 88, and 12\dfrac{1}{2} instead of 116\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

Intuition

sin⁡x/x→1\sin x/x \to 1 is a radian statement. A degree is a much smaller angle — π180\dfrac{\pi}{180} of a radian — so sin⁡x∘\sin x^\circ is roughly πx180\dfrac{\pi x}{180}, and every standard limit picks up that factor.

Definition

  • x∘=πx180x^\circ = \dfrac{\pi x}{180} radians. Convert inside every trigonometric function before using any standard limit.
  • lim⁡x→0sin⁡x∘x=π180\lim_{x\to 0}\dfrac{\sin x^\circ}{x} = \dfrac{\pi}{180}, lim⁡x→0tan⁡x∘x=π180\lim_{x\to 0}\dfrac{\tan x^\circ}{x} = \dfrac{\pi}{180}, lim⁡x→01−cos⁡x∘x2=12(π180)2\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: cos⁡mx∘−cos⁡nx∘x2→(n2−m2)π22⋅1802\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 π\pi and a power of 180180 (or a divisor of it); an answer without π\pi has ignored the degree sign.

Degree conversion in a limit

x∘=πx180 radlim⁡x→0sin⁡x∘x=π180lim⁡x→01−cos⁡x∘x2=π22⋅1802x^\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→0sin⁡x∘x\lim_{x\to 0}\dfrac{\sin x^\circ}{x}.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 5LimitsHARD
lim⁡x→0cos⁡7x∘−cos⁡2x∘x2\displaystyle\lim_{x\to0}\frac{\cos7x^{\circ}-\cos2x^{\circ}}{x^{2}} is

[Q144 · 12th May Shift 1 · 2024]

Dropping the degree sign

cos⁡7x∘−cos⁡2x∘x2\dfrac{\cos 7x^\circ - \cos 2x^\circ}{x^2} is −π21440-\dfrac{\pi^2}{1440}; the radian version would be −452-\dfrac{45}{2}, which is the first distractor. If the stem prints a small circle, the answer has a π\pi in it.

Concept 6 of 6

Higher-Order Forms: Expand to the Needed Power

Intuition

When the first-order pieces cancel — tan⁡x−sin⁡x\tan x - \sin x, xtan⁡2x−2xtan⁡xx\tan 2x - 2x\tan x — the limit lives in the x3x^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−x36+…\sin x = x - \dfrac{x^3}{6} + \dots, tan⁡x=x+x33+…\tan x = x + \dfrac{x^3}{3} + \dots, cos⁡x=1−x22+x424−…\cos x = 1 - \dfrac{x^2}{2} + \dfrac{x^4}{24} - \dots, ex=1+x+x22+…e^x = 1 + x + \dfrac{x^2}{2} + \dots, log⁡(1+x)=x−x22+…\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 x3x^3 needs cubic terms; x4x^4 needs quartic terms.
  • Scaled arguments scale the series: tan⁡2x=2x+(2x)33=2x+8x33\tan 2x = 2x + \dfrac{(2x)^3}{3} = 2x + \dfrac{8x^3}{3}.
  • Differences of standard forms: tan⁡x−sin⁡x∼x32\tan x - \sin x \sim \dfrac{x^3}{2}, x−sin⁡x∼x36x - \sin x \sim \dfrac{x^3}{6}, tan⁡x−x∼x33\tan x - x \sim \dfrac{x^3}{3}, ex−1−x∼x22e^x - 1 - x \sim \dfrac{x^2}{2}.
  • Without series: tan⁡h−sin⁡h=tan⁡h (1−cos⁡h)\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−x36+…tan⁡x=x+x33+…cos⁡x=1−x22+x424−…ex=1+x+x22+…\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→0tan⁡x−sin⁡xx3\lim_{x\to 0}\dfrac{\tan x - \sin x}{x^3}.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 6LimitsHARD
lim⁡x→0xtan⁡2x−2xtan⁡x(1−cos⁡2x)2\lim_{x\to 0}\frac{x\tan 2x-2x\tan x}{(1-\cos 2x)^2} is

[Q111 · 11th May Shift 2 · 2023]

Stopping at first order and getting 0/0 again

xtan⁡2x−2xtan⁡x(1−cos⁡2x)2\dfrac{x\tan 2x - 2x\tan x}{(1 - \cos 2x)^2}: first order gives 2x2−2x2=02x^2 - 2x^2 = 0 on top. Go to third order — x(2x+8x33)−2x(x+x33)=2x4x\left(2x + \frac{8x^3}{3}\right) - 2x\left(x + \frac{x^3}{3}\right) = 2x^4 — against (2x2)2=4x4(2x^2)^2 = 4x^4 below, and the limit is 12\dfrac{1}{2}. The cancellation is the signal to expand further, not to answer 00.

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→0sin⁡xx=1lim⁡x→0tan⁡xx=1lim⁡x→0sin⁡kxx=klim⁡x→0sin⁡axsin⁡bx=ab\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=2sin⁡2x2lim⁡x→01−cos⁡xx2=12lim⁡x→01−cos⁡kxx2=k221 - \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⁡(π−θ)=sin⁡θcos⁡A−cos⁡B=−2sin⁡A+B2sin⁡A−B22−2cos⁡ϕ=2∣sin⁡ϕ2∣\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=π2−h:sin⁡x=cos⁡h,  cos⁡x=sin⁡h,  cot⁡x=tan⁡h,  π−2x=2h,  1−sin⁡x=1−cos⁡h∼h22x = \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∘=πx180 radlim⁡x→0sin⁡x∘x=π180lim⁡x→01−cos⁡x∘x2=π22⋅1802x^\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−x36+…tan⁡x=x+x33+…cos⁡x=1−x22+x424−…ex=1+x+x22+…\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)

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