Limits — MHT-CET Maths Formulas and Traps

Limits — Existence, One-Sided Limits and Limits at Infinity

What a Limit Says

Algebra of limits

\lim (f \pm g) = \lim f \pm \lim g \qquad \lim (fg) = \lim f \cdot \lim g \qquad \lim \frac{f}{g} = \frac{\lim f}{\lim g}\ \ (\lim g \neq 0)

$\lim$ all limits taken as $x \to a$ , each assumed to exist

One-Sided Limits and When a Limit Exists

Existence of a limit

\lim_{x\to a} f(x) = L \iff \lim_{x\to a^-} f(x) = \lim_{x\to a^+} f(x) = L

Greatest-Integer and Sign Functions Near a Point

Greatest integer at an integer

\lim_{x\to n^-}[x] = n - 1, \qquad \lim_{x\to n^+}[x] = n \qquad (n \in \mathbb{Z})

Limits at Infinity — Compare the Leading Powers

Ratio of polynomials at infinity

\lim_{x\to\infty}\frac{a_n x^n + \dots + a_0}{b_m x^m + \dots + b_0} = \begin{cases} 0, & n < m \\[2pt] \dfrac{a_n}{b_m}, & n = m \\[2pt] \pm\infty, & n > m \end{cases}

$n, m$ degrees of numerator and denominator
$a_n, b_m$ their leading coefficients

A Finite Limit at Infinity Forces the Divergent Part to Vanish

Finite limit at infinity

\frac{p(x)}{q(x)} = \alpha x + \beta + \frac{r(x)}{q(x)} \ \Rightarrow\ \lim_{x\to\infty}\left(\frac{p(x)}{q(x)} - ax - b\right) \text{ finite} \iff a = \alpha,\ \text{and the limit is } \beta - b

Infinity Minus Infinity — Rationalise at Infinity

Root minus its leading term

\lim_{x\to\infty}\left(\sqrt{x^2 + ax + b} - x\right) = \frac{a}{2}

Common traps

The value at the point is not the limit

f(x) = \dfrac{x^2}{x}

has no value at

x = 0

, yet

\lim_{x\to 0} f(x) = 0

. Conversely a function can be defined at a point with a value that has nothing to do with its limit there. Keep the two questions separate — continuity is precisely the case where they agree.

Not every modulus makes the limit fail

\dfrac{x}{|x| + x^2}

tends to

+1

from the right and

-1

from the left — no limit. But

\dfrac{|x|}{|x| + x^2}

tends to

1

from BOTH sides, because the sign that flips upstairs also flips downstairs. Both versions have been set in the same paper series; the answer is decided by working the two sides, not by spotting a modulus.

[x] for a small negative x is −1, not 0

[-0.001] = -1

. Writing

0

here turns a

-\sin 1

answer into

0

, which is always one of the options. On the left of any integer

n

,

[x]

is

n - 1

— including

n = 0

.

A sum of n terms is not 'n copies of the biggest term'

\dfrac{1 + 8 + 27 + \dots + n^3}{1 - n^4}

is NOT

\dfrac{n \cdot n^3}{-n^4} = -1

. The sum is

\left(\frac{n(n+1)}{2}\right)^2 \sim \frac{n^4}{4}

, so the limit is

-\dfrac{1}{4}

. Always write the closed form of a sum before comparing powers.

Solving for b before a

There is no equation for

b

until

a

has killed the growing term. Students who 'compare constants' first get

b

from the wrong expression. Order: divide → set the

x

-coefficient to zero → read off the constant.

Subtracting infinities term by term

Writing

\sqrt{x^2 + 4x} - x \approx x - x = 0

is wrong; the answer is

2

. The lower-order terms under the root are exactly what survives after the leading parts cancel, so they cannot be dropped before rationalising.

Algebraic Limits — Factorisation, Rationalisation and the xⁿ − aⁿ Form

Learn this subtopic in the notes

Factor and Cancel

Factorisations that unlock 0/0

x^2 - a^2 = (x - a)(x + a) \qquad x^3 - a^3 = (x - a)(x^2 + ax + a^2) \qquad p(a) = 0 \Rightarrow (x - a) \mid p(x)

Rationalisation — Single, Double and Nested Surds

Conjugate rule

\frac{\sqrt{A} - \sqrt{B}}{C} = \frac{A - B}{C\left(\sqrt{A} + \sqrt{B}\right)}

The xⁿ − aⁿ Standard Form and Fractional Powers

The xⁿ − aⁿ family

\lim_{x\to a}\frac{x^n - a^n}{x - a} = n a^{n-1} \quad (n \in \mathbb{Q}) \qquad \lim_{x\to 0}\frac{(1 + x)^n - 1}{x} = n

The Derivative in Disguise — [f(x) − f(a)]/(x − a) and L'Hôpital

Derivative form and L'Hôpital

\lim_{x\to a}\frac{f(x) - f(a)}{x - a} = f'(a) \qquad \lim_{x\to a}\frac{f(x)}{g(x)} \overset{0/0}{=} \lim_{x\to a}\frac{f'(x)}{g'(x)}

A Finite Limit Forces the Numerator to Vanish — Finding a and b

Finite limit at a zero of the denominator

\lim_{x\to a}\frac{p(x)}{x - a} = L \text{ (finite)} \ \Rightarrow\ p(a) = 0 \ \text{ and } \ p'(a) = L

Common traps

Cancelling before checking the form

Cancelling

(x - a)

is only valid when it is genuinely a factor of both floors. If substitution gives

\dfrac{5}{0}

there is nothing to cancel and no finite limit — writing an option like '5' or '0' there is the standard distractor.

Two blowing-up fractions must be combined first

\dfrac{1}{x - 2} - \dfrac{2x}{x^3 - 3x^2 + 2x}

at

x \to 2

: each piece alone is

\infty

, and '

\infty - \infty = 0

' is a trap. Factor the second denominator as

x(x - 1)(x - 2)

, put everything over it, and the combined numerator

(x - 2)(x + 1)

cancels to give

\dfrac{3}{2}

.

Rationalising only one floor when both carry surds

In

\dfrac{\sqrt{a + 2x} - \sqrt{3x}}{\sqrt{3a + x} - 2\sqrt{x}}

the top rationalises to

a - x

and the bottom to

3(a - x)

; stopping after one of them leaves a

0/0

you cannot substitute into. Do both, cancel

(a - x)

, then evaluate the two conjugate factors at

x = a

to get

\dfrac{2}{3\sqrt3}

.

The inner function's sign

In

\dfrac{(84 - x)^{1/4} - 3}{x - 3}

the inner variable is

84 - x

, which DECREASES as

x

increases. Writing

u = 84 - x

gives

x - 3 = -(u - 81)

, so the answer is

-\dfrac{1}{4}\cdot 81^{-3/4} = -\dfrac{1}{108}

, not

+\dfrac{1}{108}

. Both signs are always in the options.

L'Hôpital on a form that is not indeterminate

\lim_{x\to 1}\dfrac{x^2 + 1}{x + 1}

is

1

by substitution; 'differentiating' gives

\dfrac{2x}{1} \to 2

, which is wrong. The rule has a precondition, and the paper's distractors are built from students who skip it.

Treating a as free and reading b off the limit

Skipping

p(a) = 0

leaves one equation for two unknowns, and every option looks reachable. The vanishing condition is not optional — it is the reason the limit is finite at all.

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

Learn this subtopic in the notes

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

Common traps

sin x/x → 1 only as x → 0

As

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.

Treating 1 − cos x as first order

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

is

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.

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

In

\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.

(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.

Dropping the degree sign

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

is

-\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.

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

.

Exponential, Logarithmic and 1^∞ Limits

Learn this subtopic in the notes

The Exponential and Logarithmic Standard Limits

Exponential and logarithmic standard limits

\lim_{x\to 0}\frac{a^x - 1}{x} = \log a \qquad \lim_{x\to 0}\frac{e^x - 1}{x} = 1 \qquad \lim_{x\to 0}\frac{\log(1 + x)}{x} = 1 \qquad \lim_{x\to 0}\frac{\log(1 + kx)}{x} = k

$\log$ natural logarithm (base $e$ )

Factorising aˣ − bˣ − cˣ + 1 into (bˣ − 1)(cˣ − 1)

Grouping factorisation

(bc)^x - b^x - c^x + 1 = (b^x - 1)(c^x - 1) \sim x^2\,\log b\,\log c

Substitute t = aˣ When the Exponents Are Mixed

Exponential substitution

t = a^{x/k}\ \Rightarrow\ a^{x} = t^{k},\quad a^{c - x} = \frac{a^{c}}{t^{k}},\qquad x \to x_0 \iff t \to a^{x_0/k}

Composite Forms: (eᵘ − 1)/u with u → 0, and Mixed Series Terms

Composite exponential forms

e^{a} - e^{b} = e^{b}\left(e^{a - b} - 1\right) \qquad \lim_{u\to 0}\frac{e^{u} - 1}{u} = 1 \ \text{ for any } u = u(x) \to 0

The 1^∞ Form: lim f^g = e^{lim (f − 1)g}

The 1^∞ rule

\lim_{x\to 0}(1 + x)^{1/x} = e \qquad \lim_{x\to\infty}\left(1 + \frac{k}{x}\right)^{x} = e^{k} \qquad f \to 1,\ g \to \infty:\ \lim f^{g} = e^{\lim (f - 1)g}

0⁰ and ∞⁰ Forms — Take Logarithms

Power forms through the logarithm

l = \lim f^{g} \ \Rightarrow\ \log l = \lim g\log f \qquad \lim_{x\to 0^+} x\log x = 0 \ \Rightarrow\ \lim_{x\to 0^+} x^{x} = 1

Common traps

(aˣ − 1)/x is log a, never a

The limit is the slope of

a^x

at

0

, which is

\log a

. An option reading

\dfrac{3}{2}

where

\log\dfrac{3}{2}

is correct is the standard distractor.

Sending each term to its own limit

63^x - 9^x - 7^x + 1 \to 1 - 1 - 1 + 1 = 0

tells you the form is

0/0

and nothing else. The value comes from the factorised product, and it is second order — pair it with

x^2

or

1 - \cos x

, not with

x

.

Choosing t too large

Putting

t = 5^x

leaves

5^{x/2} = \sqrt{t}

, a surd that needs rationalising. Choosing the SMALLEST power as

t

keeps everything polynomial.

Using only first order on eˣ² − cos x

\dfrac{e^{x^2} - \cos x}{x^2}

needs the

x^2

term of BOTH series:

x^2

from

e^{x^2}

and

\dfrac{x^2}{2}

from

\cos x

, giving

\dfrac{3}{2}

. Dropping the cosine's

\frac12

gives

1

, which is among the options.

1^∞ is indeterminate — never answer 1 by substitution

\left(\dfrac{x + 8}{x + 1}\right)^{x + 5}

looks like

1^\infty = 1

and equals

e^{7}

. The option '1' is present precisely for the student who substitutes.

Treating (sin x)^(1/x) as a 0⁰ form

The exponent

1/x \to +\infty

, not

0

. A base below

1

raised to an unbounded power is

0

; no logarithm is needed. Only

0^0

,

\infty^0

and

1^\infty

are indeterminate among the power forms.

Continuity at a Point — Finding f(c) and the Parameter

Learn this subtopic in the notes

Continuity at a Point — The Three-Part Test

Continuity at c

f \text{ continuous at } c \iff \lim_{x\to c^-} f(x) = \lim_{x\to c^+} f(x) = f(c)

Removable Discontinuity: Define f(c) as the Limit

Filling a removable discontinuity

f \text{ continuous at } c,\ f \text{ undefined at } c \text{ by formula} \ \Rightarrow\ f(c) := \lim_{x\to c} f(x)

Products of Standard Forms in Continuity Dress

Order bookkeeping

\frac{\prod (\text{factors} \sim c_i x^{p_i})}{\prod (\text{factors} \sim d_j x^{q_j})} \to \frac{\prod c_i}{\prod d_j} \ \text{ exactly when } \sum p_i = \sum q_j

Continuity at a Non-Zero Point: Shift to h → 0

Two routes at a non-zero point

x = c + h \ (h \to 0) \qquad \text{or} \qquad \lim_{x\to c}\frac{p(x)}{q(x)} \overset{0/0}{=} \frac{p'(c)}{q'(c)} \ \text{ when } q'(c) \neq 0

1^∞ in Continuity Problems: k = e^{…}

The 1^∞ continuity value

f(c) = \lim_{x\to c} u(x)^{v(x)} = e^{\lim_{x\to c}(u - 1)\,v} \qquad (u \to 1,\ v \to \infty)

The Parameter Inside the Function

Parameter through a standard limit

\lim_{x\to 0}\frac{\cos ax - \cos bx}{\cos cx - \cos bx} = \frac{b^2 - a^2}{b^2 - c^2} \qquad \frac{d}{dx}\int_{c}^{u(x)}\phi(t)\,dt = \phi(u(x))\,u'(x)

Common traps

A limit existing is not continuity

\dfrac{x^2 - 4}{x - 2}

has a perfectly good limit at

2

; it is still discontinuous there if

f(2)

is undefined or set to the wrong number. The test has three parts, and the third one is where the marks are.

Computing f(c) from the formula

Substituting

x = c

into the formula divides by zero — that is the whole reason the point is special.

f(c)

is the LIMIT, obtained by resolving the

0/0

.

Powers that do not match

If the numerator is order

x^2

and the denominator order

x^3

, the limit is infinite and no

k

makes

f

continuous — the answer to 'find

k

' would be 'no such

k

'. When your bookkeeping gives that on a paper that offers four numbers, re-read the stem: a lost square or a misread

f(0)

is the usual cause.

Applying the quotient rule instead of L'Hôpital

L'Hôpital differentiates the top and the bottom SEPARATELY. Differentiating the fraction as a whole is a different operation and gives a different, wrong number.

A fixed base is not 1^∞

\left(\frac45\right)^{\tan 4x/\tan 5x}

at

x \to \frac{\pi}{2}

: the exponent is

-\tan 4h\tan 5h \to 0

, so the power is

\left(\frac45\right)^{0} = 1

and

k + \frac25 = 1

gives

k = \frac35

. Reaching for

e^{\dots}

here is the wrong tool.

Differentiating the integral without the chain factor

(dfrac{d}{dx}int_{3}^{f(x)} 3t^2,dt = 3[f(x)]^2cdot f'(x)). Dropping (f'(x)) gives (3cdot 9 = 27); with it, (27cdotrac{1}{27} = 1). The chain factor is the whole question.

Continuity of Piecewise Functions — Junction Conditions and Parameter Systems

Learn this subtopic in the notes

One Junction: Left Limit = Right Limit = Value

Junction condition

f(x) = \begin{cases} g(x), & x \le c \\ h(x), & x > c \end{cases} \text{ continuous at } c \iff g(c) = \lim_{x\to c^+} h(x)

Two Different Formulas Meeting at 0: Compute Each Side with Its Own Tool

The two recurring one-sided limits

\lim_{x\to 0^-}\frac{1 - \cos 4x}{x^2} = 8 \qquad \lim_{x\to 0^+}\frac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4} = \lim_{x\to 0^+}\left(\sqrt{16 + \sqrt{x}} + 4\right) = 8

Exponential Junctions and the Given Value at 0

Exponential-denominator junction

\lim_{x\to 0}\frac{p\sin x + q\tan x}{a^{x} - 1} = \frac{p + q}{\log a} \qquad \log a = k\log 2 \iff a = 2^{k}

Two Junctions, Two Unknowns: Set Up a Linear System

The two-seam system

\text{seam } c_1:\ g(c_1) = h(c_1) \qquad \text{seam } c_2:\ h(c_2) = k(c_2) \qquad \Rightarrow\ \text{two linear equations in } a, b

The Squeeze: x² sin(1/x) Is Continuous for Any Coefficient

Squeeze at 0

\left|x^2\sin\frac{1}{x}\right| \le x^2 \to 0 \ \Rightarrow\ \lim_{x\to 0} x^2\sin\frac{1}{x} = 0

Common traps

Which piece owns the point?

f(c)

comes from the piece whose inequality includes

c

— the one written with

\le

or

\ge

. When a stem gives a separate value at

c

(

f(3) = a + b

), that value is a THIRD quantity and must equal both one-sided limits.

Substituting into the surd piece

\dfrac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4}

is

\dfrac{0}{0}

at

0

, not

0

. Multiply by the conjugate

\sqrt{16 + \sqrt{x}} + 4

; the

\sqrt{x}

cancels and the value is

8

. Answering

0

or

-8

(a sign slip in the conjugate) are the two distractors.

log a = 4 log 2 means a = 16, not a = 8

4\log 2 = \log 2^4 = \log 16

. The option

a = 8

comes from

2 \times 4

; the exponent rule for logs is what the stem is testing.

Applying continuity at only one seam

With two unknowns, one equation leaves a free parameter, and any option can be 'reached'. The 2022 sitting's own key was wrong for exactly this reason — it used one seam and printed

\frac{13}{5}

where two seams give

\frac{23}{5}

. Count the seams before solving.

Looking for a constraint that is not there

When both pieces tend to

0

for every value of the parameters, the answer is 'any real

a

, any real

b

'. Options that restrict

a

to rationals or irrationals are noise — nothing in the limit distinguishes them.

Discontinuities of [x], |x| and sgn x — Counting the Points

Learn this subtopic in the notes

[x] Is Discontinuous at Every Integer — Counting Them

Jumps of the greatest-integer function

\lim_{x\to n^-}[x] = n - 1 \ne n = [n] \quad (n \in \mathbb{Z}) \qquad [kx] \text{ jumps at } x = \frac{m}{k},\ m \in \mathbb{Z}

Signum-Type Jumps: (x − a)/|x − a| and Products with It

The sign factor

\frac{x - a}{|x - a|} = \begin{cases} 1, & x > a \\ -1, & x < a \end{cases} \qquad \text{jump of size } 2 \text{ at } a

Piecewise with [x] and |x|: Check Every Join and Both Endpoints

Continuity at a closed endpoint

\text{left endpoint } a:\ \lim_{x\to a^+} f(x) = f(a) \qquad \text{right endpoint } b:\ \lim_{x\to b^-} f(x) = f(b)

Composites of [x]: Where Does the Inner Function Cross an Integer?

Jumps of a composite

[g(x)] \text{ jumps where } g(x) \in \mathbb{Z} \text{ and } g \text{ crosses it} \qquad [-u] = -[u] - 1 \ (u \notin \mathbb{Z})

When the Other Factor Vanishes at the Jump

A zero swallows a bounded jump

g \text{ continuous at } n,\ g(n) = 0,\ |h| \le M \text{ near } n \ \Rightarrow\ \lim_{x\to n} g(x)h(x) = 0 = g(n)h(n)

Common traps

Miscounting the negative side

In

\left(-\frac{7}{2}, 100\right)

the integers are

-3, -2, -1

(three of them), then

0

, then

1

to

99

. Forgetting

0

, or including

100

, gives

102

or

104

— both offered.

Cancelling |x − 1| against (x − 1)

\dfrac{x - 1}{|x - 1|}

is not

1

; it is

\pm 1

depending on the side. Cancelling as if the modulus were absent loses the jump and produces 'continuous on

\mathbb{R}

', which is always an option.

Forgetting the closed right endpoint

On

[-1, 3]

with

f(x) = x + [x]

near

3

, the value

f(3) = 6

is not the left limit

5

. A student who only checks interior seams reports two points; the answer is three.

Answering the 'find k' question when no k exists

[x^2] - [-x^2]

at

3

has one-sided limits

17

and

19

. The exam key takes the right-hand value

19

; mathematically no

k

works. On the paper choose

19

; in your notes, know why the question is flawed.

The exam key that contradicts the mathematics

For

[x]\cos\dfrac{(2x - 1)\pi}{2}

the official MHT-CET key marks 'all integer points', reasoning only that

[x]

jumps there. The cosine is

0

at every integer, so the function is in fact continuous everywhere. Know the correct mathematics; on that particular paper, the marked answer was the key's.

Limits formulas

Limits — Existence, One-Sided Limits and Limits at Infinity

What a Limit Says

One-Sided Limits and When a Limit Exists

Greatest-Integer and Sign Functions Near a Point

Limits at Infinity — Compare the Leading Powers

A Finite Limit at Infinity Forces the Divergent Part to Vanish

Infinity Minus Infinity — Rationalise at Infinity

Common traps

Algebraic Limits — Factorisation, Rationalisation and the xⁿ − aⁿ Form

Factor and Cancel

Rationalisation — Single, Double and Nested Surds

The xⁿ − aⁿ Standard Form and Fractional Powers

The Derivative in Disguise — [f(x) − f(a)]/(x − a) and L'Hôpital

A Finite Limit Forces the Numerator to Vanish — Finding a and b

Common traps

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

sin x/x, tan x/x and Scaled Arguments

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

Rewrite with an Identity Before Taking the Limit

Shift the Variable: Limits at π/2 and Other Non-Zero Points

Degrees Are Not Radians

Higher-Order Forms: Expand to the Needed Power

Common traps

Exponential, Logarithmic and 1^∞ Limits

The Exponential and Logarithmic Standard Limits

Factorising aˣ − bˣ − cˣ + 1 into (bˣ − 1)(cˣ − 1)

Substitute t = aˣ When the Exponents Are Mixed

Composite Forms: (eᵘ − 1)/u with u → 0, and Mixed Series Terms

The 1^∞ Form: lim f^g = e^{lim (f − 1)g}

0⁰ and ∞⁰ Forms — Take Logarithms

Common traps

Continuity at a Point — Finding f(c) and the Parameter

Continuity at a Point — The Three-Part Test

Removable Discontinuity: Define f(c) as the Limit

Products of Standard Forms in Continuity Dress

Continuity at a Non-Zero Point: Shift to h → 0

1^∞ in Continuity Problems: k = e^{…}

The Parameter Inside the Function

Common traps

Continuity of Piecewise Functions — Junction Conditions and Parameter Systems

One Junction: Left Limit = Right Limit = Value

Two Different Formulas Meeting at 0: Compute Each Side with Its Own Tool

Exponential Junctions and the Given Value at 0

Two Junctions, Two Unknowns: Set Up a Linear System

The Squeeze: x² sin(1/x) Is Continuous for Any Coefficient

Common traps

Discontinuities of [x], |x| and sgn x — Counting the Points

[x] Is Discontinuous at Every Integer — Counting Them

Signum-Type Jumps: (x − a)/|x − a| and Products with It

Piecewise with [x] and |x|: Check Every Join and Both Endpoints

Composites of [x]: Where Does the Inner Function Cross an Integer?

When the Other Factor Vanishes at the Jump

Common traps

More MHT-CET Maths formula sheets