MHT-CET Maths · Limits

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

f is continuous at c when the limit exists and equals f(c) — so a 'find k' or 'find f(0)' question is a limit from the earlier pages, set equal to a value.

All Limits notes MHT-CET Maths strategy

Why this matters

34 PYQs at 68% HARD, the largest page in the chapter and pure recycling: every question here is a limit from the four pages before it, evaluated and then equated. The extra difficulty is clerical — a parameter buried inside the limit, a point that is not 0, an integral in the numerator — never a new idea. Four of these stems were set twice in different sittings with identical numbers, so the recurring forms are worth recognising on sight.

Concept 1 of 6: Continuity at a Point — The Three-Part Test

A function is continuous at

c

if you can draw it through

c

without lifting the pen: the left approach, the right approach and the actual dot at

c

all agree.

Definition

$f$ is continuous at $c$ when three things hold: $f(c)$ exists; $\lim_{x\to c} f(x)$ exists (left = right); and the two are equal.
Removable discontinuity: the limit exists but $f(c)$ is missing or different — a hole, fixable by redefining one value. Jump: left and right limits differ. Infinite: the function blows up.
Polynomials, $\sin$ , $\cos$ , $e^x$ are continuous everywhere; rational functions, $\tan$ , $\log$ , roots are continuous at every point of their domain. So discontinuity can only happen where a formula changes or a denominator vanishes.
Every 'is continuous at $c$ , find $k$ ' stem is the equation $\lim_{x\to c} f(x) = f(c)$ with $k$ on one side.
Continuity on an interval means continuity at every interior point plus the appropriate one-sided continuity at the endpoints.

Continuity at c

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

Worked example

f(x) = \dfrac{x^2 - 4}{x - 2}

for

x \neq 2

and

f(2) = 5

. Is

f

continuous at

2

? If not, how should

f(2)

be redefined?

Practise this concept4 quick reps

The same idea in a real exam question:

MHT-CET · 2021 · May Shift 1 · Q118Moderate

Example 1 · Limits · Continuity at a Point — Finding f(c) and the Parameter

f(x) = \begin{cases}\frac{3\sin(\pi x)}{5x} & x \neq 0 \\ 2k & x = 0\end{cases}

is continuous at

x = 0

, then the value of

k

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.

Concept 2 of 6: Removable Discontinuity: Define f(c) as the Limit

When the formula is undefined at exactly one point, continuity has only one possible meaning:

f(c)

must be whatever the formula is heading towards. The question 'find

f(c)

so that

f

is continuous' is the question 'evaluate the limit'.

Definition

Template: $f(x) = \dfrac{\text{something}}{\text{something that vanishes at } c}$ for $x \neq c$ , continuous at $c$ . Then $f(c) = \lim_{x\to c} f(x)$ , full stop.
The limit is one of the earlier pages: factor, rationalise, $x^n - a^n$ , a trigonometric or exponential standard form.
Cube and fifth roots at $0$ : $(27 - 2x)^{1/3} - 3$ and $9 - 3(243 + 5x)^{1/5}$ are both derivative-in-disguise forms — differentiate top and bottom once (L'Hôpital) and substitute $x = 0$ .
A sum $x + x^2 + \dots + x^n - n$ over $x - 1$ splits into $\dfrac{x - 1}{x - 1} + \dfrac{x^2 - 1}{x - 1} + \dots$ , giving $1 + 2 + \dots + n$ .
Trigonometric points other than $0$ ( $x \to 1$ with $\cos\pi x$ , $x \to \pi/4$ ): shift the variable as on the trigonometric page.

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)

Worked example

f(x) = \dfrac{\sqrt{x + 4} - 2}{x}

for

x \neq 0

is continuous at

0

. Find

f(0)

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 2nd May Shift 1 · Q110Hard

Example 2 · Limits · Continuity at a Point — Finding f(c) and the Parameter

f(x) = \dfrac{1 + \cos(\pi x)}{\pi(1-x)^2}

for

x \ne 1

is continuous at

x = 1

, then

f(1)

is equal to

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

Drill 4 more on removable discontinuity: define f(c) as the limit

Concept 3 of 6: Products of Standard Forms in Continuity Dress

The HARD continuity stems stack three or four standard forms into one fraction —

\sin^3\sqrt{x}

over

(\tan^{-1}\sqrt{x})^2

\log(1 + 3x)

over

e^{5\sqrt{x}} - 1

. Replace each by its order, multiply the powers of

x

, and the fraction collapses to a constant.

Definition

Replace every factor by its leading behaviour: $\sin u \to u$ , $\tan u \to u$ , $\tan^{-1}u \to u$ , $e^{u} - 1 \to u$ , $a^{u} - 1 \to u\log a$ , $\log(1 + u) \to u$ , $1 - \cos u \to \dfrac{u^2}{2}$ , each valid because $u \to 0$ .
Then total the powers of $x$ upstairs and downstairs, treating $\sqrt{x}$ as $x^{1/2}$ . Equal totals → the constant that remains is $f(c)$ . Unequal → the limit is $0$ or infinite and no value of the parameter rescues continuity.
Exponential differences factor first: $10^x + 7^x - 14^x - 5^x = (2^x - 1)(5^x - 7^x)$ , each bracket first order, so the product pairs with $1 - \cos x$ .
Keep $\log a - \log b = \log\dfrac{a}{b}$ and $2\log 2 = \log 4$ ready: the option list is written in a single log.

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

Worked example

f(x) = \dfrac{(e^{x} - 1)\sin 2x}{x\log(1 + 3x)}

for

x \neq 0

is continuous at

0

. Find

f(0)

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 23 April Shift I · Q109Hard

Example 3 · Limits · Continuity at a Point — Finding f(c) and the Parameter

f(x) =\frac{10^{x}+7^{x}-14^{x}-5^{x}}{1 - \cos x},x\neq 0

is continuous at

x= 0

, then the value of

f(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.

Drill 3 more on products of standard forms in continuity dress

Concept 4 of 6: Continuity at a Non-Zero Point: Shift to h → 0

x = \dfrac{\pi}{4}

\dfrac{\pi}{2}

the standard limits do not apply directly. Either shift the variable so the point becomes

0

, or — faster when both floors are differentiable — read the ratio as a derivative and apply L'Hôpital once.

Definition

Shift: at $c = \dfrac{\pi}{2}$ put $x = \dfrac{\pi}{2} - h$ ; at $c = \dfrac{\pi}{4}$ put $x = \dfrac{\pi}{4} + h$ and use $\tan\left(\dfrac{\pi}{4} + h\right) = \dfrac{1 + \tan h}{1 - \tan h}$ .
L'Hôpital: for $\dfrac{1 - \tan x}{4x - \pi}$ at $\dfrac{\pi}{4}$ , differentiate: $\dfrac{-\sec^2 x}{4} \to \dfrac{-2}{4} = -\dfrac{1}{2}$ . For $\dfrac{\sqrt2\cos x - 1}{\cot x - 1}$ : $\dfrac{-\sqrt2\sin x}{-\csc^2 x} \to \dfrac{-1}{-2} = \dfrac{1}{2}$ .
Both routes must agree; use the shift when a power of $(\pi - 2x)$ sits below (L'Hôpital would need repeating), and L'Hôpital when the denominator is linear in $x$ .
The value $f(c)$ or $k$ is then the number obtained — nothing else changes.

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

Worked example

f(x) = \dfrac{\cos x}{\pi - 2x}

for

x \neq \dfrac{\pi}{2}

is continuous at

\dfrac{\pi}{2}

. Find

f\!\left(\dfrac{\pi}{2}\right)

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 3rd May Shift 1 · Q103Moderate

Example 4 · Limits · Continuity at a Point — Finding f(c) and the Parameter

Let

f(x)=\frac{1-\tan x}{4x-\pi},\ x\neq\frac{\pi}{4},\ x\in\!\left[0,\frac{\pi}{2}\right]

. If

f(x)

is continuous in

\left[0,\frac{\pi}{2}\right]

, then

f\!\left(\frac{\pi}{4}\right)

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.

Drill 2 more on continuity at a non-zero point: shift to h → 0

Concept 5 of 6: 1^∞ in Continuity Problems: k = e^{…}

When the formula is a power whose base

\to 1

and exponent

\to \infty

at the point, the value that makes

f

continuous is

e

to a limit — and the option list is written as

e^{6}, e^{2}, e^{-6}, e^{-2}

Definition

$k = f(c) = \lim f^{g} = e^{\lim (f - 1)g}$ , the rule from the exponential page.
Rational base: $\left(\dfrac{5x - 8}{8 - 3x}\right)^{3/(2x - 4)}$ at $x \to 2$ : base $\to 1$ , exponent $\to \infty$ ; $f - 1 = \dfrac{8x - 16}{8 - 3x}$ , and $(f - 1)g = \dfrac{3(8x - 16)}{(8 - 3x)(2x - 4)} = \dfrac{24}{2(8 - 3x)} \to \dfrac{24}{4} = 6$ , so $k = e^{6}$ .
Trigonometric base: $\left(\dfrac{1 + \tan x}{1 + \sin x}\right)^{\csc x}$ at $0$ : $f - 1 = \dfrac{\tan x - \sin x}{1 + \sin x} \sim \dfrac{x^3}{2}$ , times $\csc x \sim \dfrac{1}{x}$ gives $\to 0$ , so $k = e^{0} = 1$ .
Not $1^\infty$ : $\left(\dfrac{4}{5}\right)^{\tan 4x/\tan 5x}$ at $\dfrac{\pi}{2}$ has a fixed base; the exponent $\to 0$ , so the limit is simply $1$ .

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)

Worked example

f(x) = (1 + 2x)^{1/x}

for

x \neq 0

is continuous at

0

. Find

f(0)

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 3rd May Shift 2 · Q104Hard

Example 5 · Limits · Continuity at a Point — Finding f(c) and the Parameter

If the function

f(x)=\left(\dfrac{5x-8}{8-3x}\right)^{\frac{3}{2x-4}}

x\neq2

=k

x=2

, is continuous at

x=2

, then

k=

A fixed base is not 1^∞

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

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.

Drill 2 more on 1^∞ in continuity problems: k = e^{…}

Concept 6 of 6: The Parameter Inside the Function

Sometimes the unknown is not

f(c)

but a constant buried inside the formula —

k

\sin\dfrac{x}{k}

, or

a, b, c

\dfrac{\cos ax - \cos bx}{\cos cx - \cos bx}

. Evaluate the limit with the parameter as a symbol, equate to the given

f(c)

, and solve.

Definition

Carry the parameter through the standard limits: $\sin\dfrac{x}{k} \sim \dfrac{x}{k}$ , $1 - \cos ax \sim \dfrac{a^2x^2}{2}$ , $\dfrac{a^{x} - 1}{x} \to \log a$ .
$\dfrac{\cos ax - \cos bx}{\cos cx - \cos bx} \to \dfrac{b^2 - a^2}{b^2 - c^2}$ ; setting this equal to $-1$ gives $a^2 + c^2 = 2b^2$ — $a^2, b^2, c^2$ in arithmetic progression.
The equation may be quadratic ( $k^2 = 16$ , $a^2 = 2$ ); the stem or the options decide the sign.
Integral in the numerator: $g(x) = \dfrac{\int_{3}^{f(x)} 3t^2\,dt}{x - 3}$ is $0/0$ at $3$ when $f(3) = 3$ ; L'Hôpital with the fundamental theorem gives $3[f(x)]^2 f'(x) \to 3\cdot 9\cdot f'(3)$ . Differentiate the integral by substituting the upper limit and multiplying by its derivative.

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)

Worked example

f(x) = \dfrac{\cos ax - \cos 3x}{x^2}

for

x \neq 0

and

f(0) = 4

is continuous at

0

. Find

a

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 26 April Shift I · Q116Hard

Example 6 · Limits · Continuity at a Point — Finding f(c) and the Parameter

If the function

f(x) =\left\{ \begin{matrix} \dfrac{\cos ax- \cos bx}{\cos cx- \cos bx} & ,\ \text{if } x\neq 0 \ -1 & ,\ \text{if } x = 0 \end{matrix} \right.

is continuous at

x = 0

, then

a^{2}, b^{2}, c^{2}

are in

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.

Drill 1 more on the parameter inside the function

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)

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

Watch out for (6)

A limit existing is not continuity
→ Continuity at a Point — The Three-Part Test
Computing f(c) from the formula
→ Removable Discontinuity: Define f(c) as the Limit
Powers that do not match
→ Products of Standard Forms in Continuity Dress
Applying the quotient rule instead of L'Hôpital
→ Continuity at a Non-Zero Point: Shift to h → 0
A fixed base is not 1^∞
→ 1^∞ in Continuity Problems: k = e^{…}
Differentiating the integral without the chain factor
→ The Parameter Inside the Function

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Related notes

Trigonometric Limits — the shift to h → 0 in full →