limx→π/2(2x-π)/(cos x) (1) 2 (2) 1 (3) -2 (4) 0 – Class 11 Mathematics | ConceptFirst

Exercise 9.6 · Q2

\displaystyle\lim_{x\to\pi/2}\dfrac{2x-\pi}{\cos x}

(1)

2

(2)

1

(3)

-2

(4)

0

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Concept understanding — Standard Limits

A Toolkit of Named Limits

Some limits recur so often across problems that it is worth memorising their values outright, along with the one theorem that proves the trickiest of them: the Sandwich (Squeeze) Theorem.

The Sandwich Theorem

Theorem 9.5. If $g(x) \le f(x) \le h(x)$ for all $x$ near $x_0$ (except possibly at $x_0$ itself), and if

\lim_{x\to x_0} g(x) = \lim_{x\to x_0} h(x) = l,

then $\lim_{x\to x_0} f(x) = l$ too — $f$ is "squeezed" between two functions that agree in the limit, so it has no room to do anything else.

Illustration: to show $\displaystyle\lim_{x\to 0} x^2\sin\!\frac{1}{x^2} = 0$ , note that $\sin(\cdot)$ is always between $-1$ and $1$ , so $-x^2 \le x^2\sin\frac{1}{x^2} \le x^2$ . Since both $-x^2$ and $x^2$ tend to $0$ as $x\to 0$ , the Sandwich Theorem forces the middle expression to $0$ as well — even though $\lim_{x\to 0}\sin\frac{1}{x^2}$ on its own does not exist (it oscillates wildly), so the product rule alone could never have been applied directly.

Watch out

This is exactly why the Sandwich Theorem is indispensable rather than a curiosity: whenever one factor oscillates without a limit but is bounded, and the other factor is squeezed to zero, the ordinary product law (Concept 2) is not applicable — you need the sandwich.

The two flagship trigonometric limits

Result 9.1.

\text{(a)}\quad \lim_{\theta\to 0}\frac{\sin\theta}{\theta} = 1 \qquad\qquad \text{(b)}\quad \lim_{\theta\to 0}\frac{1-\cos\theta}{\theta} = 0

Part (a) is proved geometrically by sandwiching $\frac{\sin\theta}{\theta}$ between $\cos\theta$ and $1$ using the areas of a triangle, a sector, and a larger triangle built on the unit circle; since both bounding functions tend to $1$ as $\theta\to0$ , so must $\frac{\sin\theta}{\theta}$ . Part (b) follows algebraically from (a) by writing $1-\cos\theta = 2\sin^2\frac{\theta}{2}$ and splitting the quotient into a $\sin$ -over-argument piece (which uses part (a)) times a factor that vanishes.

A direct corollary worth keeping separate: $\displaystyle\lim_{x\to 0}\sin x = 0$ , obtained from the sandwich $-|x|\le \sin x \le |x|$ .

The full standard-limit toolkit (§9.2.10)

Alongside the trig pair above, these are worth having on instant recall — none require anything beyond algebra and substitution to use (their proofs, where given, lean on the exponential/log relationship or on the trig pair):

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

\lim_{x\to 0}\frac{\sin^{-1}x}{x}=1 \qquad\quad \lim_{x\to 0}\frac{\tan^{-1}x}{x}=1

And the three equivalent forms of the number $e$ as a limit:

\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x = e \qquad\quad \lim_{x\to 0}(1+x)^{1/x} = e \qquad\quad \lim_{x\to\infty}\left(1+\frac{k}{x}\right)^x = e^k

Note

$e$ is a transcendental number — it never satisfies any polynomial equation with rational coefficients. That's part of why it shows up as a genuinely new limiting constant here rather than something expressible in simpler closed form.

The recognise-and-substitute pattern

Nearly every "hard-looking" limit in this section is really one of the above standard forms in disguise, reached via a clean substitution $y = (\text{some expression in } x)$ chosen so that $y \to 0$ (or $y\to\infty$ ) exactly when $x$ does. The book's worked examples all follow this shape:

Spot the shell. Identify which standard form the expression resembles — a $\left(1+\Box\right)^{1/\Box}$ shape signals $e$ ; a $\dfrac{\sin(\Box)}{\Box}$ shape signals Result 9.1(a); a $\dfrac{a^{\Box}-1}{\Box}$ shape signals Result 9.3.
Substitute $y$ for the " $\Box$ " so the expression matches the standard form exactly, tracking what $y \to$ as $x \to x_0$ .
Apply the standard limit to the $y$ -expression, then (if the exponent or coefficient outside doesn't vanish) combine using the power/product rules from Concept 2.

Fresh illustration: $\displaystyle\lim_{x\to 0} (1+3x)^{2/x}$ . Write $(1+3x)^{2/x} = \left[(1+3x)^{1/3x}\right]^{6}$ . As $x\to 0$ , let $y = 3x \to 0$ , so the inner bracket $\to e$ by the standard form $(1+y)^{1/y}\to e$ . Hence the whole limit is $e^6$ .

A second flavour — a compound quotient of trig limits: $\displaystyle\lim_{x\to 0}\frac{\sin \alpha x}{\sin \beta x} = \lim_{x\to0}\left(\frac{\sin \alpha x}{\alpha x}\cdot \frac{\alpha}{\beta}\cdot\frac{\beta x}{\sin \beta x}\right) = 1\cdot\frac{\alpha}{\beta}\cdot 1 = \frac{\alpha}{\beta}$ — each piece separately matched to Result 9.1(a) via its own substitution.

Tip

When a limit mixes several standard forms multiplicatively (e.g. a $\sin$ -quotient times an exponential-quotient), split it into a product of separate limits first (valid whenever each piece's limit exists, by the product law in Concept 2), match each piece to its own standard form independently, then multiply the results.

Watch out

These standard limits are only valid in the exact $\to 0$ (or $\to\infty$ ) shell shown. $\dfrac{\sin\theta}{\theta}\to 1$ is true only as $\theta \to 0$ — plugging in a $\theta$ that merely looks small, or forgetting to substitute so that the "argument" and the "denominator" are identical, is the single most common error in applying this toolkit.

Topics like "standard limits formula list class 11" and "sandwich theorem important questions" are commonly searched by students covering the Limits and Derivatives chapter of the NCERT-aligned CBSE Class 11 Mathematics syllabus, since this toolkit of named limits is tested every year in board exams and forms the backbone of limit-evaluation questions in JEE Main and JEE Advanced. Memorising the trigonometric, exponential and $e$ -related limits together, rather than in isolation, is what makes the "recognise-and-substitute" technique fast under exam time pressure.

Step 1. Substitute $x=\dfrac\pi2+h$ so that $h\to0$ as $x\to\dfrac\pi2$ .

Step 2. $2x-\pi = 2\left(\dfrac\pi2+h\right)-\pi = 2h$ .

Step 3. $\cos x=\cos\left(\dfrac\pi2+h\right) = \cos\dfrac\pi2\cos h-\sin\dfrac\pi2\sin h = -\sin h$ .

Step 4. So $\dfrac{2x-\pi}{\cos x}=\dfrac{2h}{-\sin h}=-2\cdot\dfrac{h}{\sin h}$ .

Step 5. As $h\to0$ , $\dfrac{h}{\sin h}\to1$ , so the limit is $-2(1)=-2$ .

✓Final answer

-2

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 31 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$\displaystyle\lim_{x \to 0} \dfrac{\sin 4x}{\sin 2x} = ?$
(a) 0
(b) 2
(c) $\dfrac{1}{2}$
(d) 1
›Reveal solutionSolution

Rewrite using the standard limit $\lim_{\theta\to0}\dfrac{\sin\theta}{\theta}=1$ applied separately to $4x$ and $2x$ .

$\lim_{x\to0}\dfrac{\sin4x}{\sin2x} = \lim_{x\to0}\dfrac{\dfrac{\sin4x}{4x}\times 4x}{\dfrac{\sin2x}{2x}\times 2x} = \lim_{x\to0}\dfrac{\sin4x}{4x}\times\dfrac{2x}{\sin2x}\times\dfrac{4x}{2x}$

As $x\to0$ , $\dfrac{\sin4x}{4x}\to1$ and $\dfrac{\sin2x}{2x}\to1$ , leaving:

$= 1\times1\times\dfrac{4x}{2x} = 2$

✓Final answer
$(b) 2.$
CBSE 2026Set ANNUAL1 markMCQ
$\displaystyle\lim_{x \to 0} \dfrac{\tan x}{x}$
(a) $0$
(b) $1$
(c) $-1$
(d) $2$
›Reveal solutionSolution

$\lim_{x\to0}\dfrac{\tan x}{x}=1$ , using the standard limit $\lim_{x\to0}\dfrac{\sin x}{x}=1$ together with $\cos 0=1$ .

Write $\tan x = \dfrac{\sin x}{\cos x}$ , so

$\dfrac{\tan x}{x} = \dfrac{\sin x}{x}\cdot\dfrac{1}{\cos x}$

As $x\to 0$ : $\dfrac{\sin x}{x}\to 1$ (the standard trigonometric limit) and $\cos x \to \cos 0 = 1$ , so $\dfrac{1}{\cos x}\to 1$ .

Therefore $\lim_{x\to0}\dfrac{\tan x}{x} = 1\times 1 = 1$ .

✓Final answer
$\displaystyle\lim_{x\to 0}\dfrac{\tan x}{x}=1$
CBSE 2026Set ANNUAL1 mark
$\displaystyle\lim_{x\to 0} \dfrac{\sin x}{x}$
›Reveal solutionSolution

$\displaystyle\lim_{x\to 0}\dfrac{\sin x}{x}=1$ is a standard limit, provable via the sandwich (squeeze) theorem using the unit circle.

For small $x$ (in radians), geometric comparison of the areas of a triangle, a circular sector, and a larger triangle bounding the unit circle gives $\sin x < x < \tan x$ for $0<x<\tfrac{\pi}{2}$ .

Dividing through by $\sin x$ and taking reciprocals: $\cos x < \dfrac{\sin x}{x} < 1$ .

As $x\to0$ , $\cos x\to1$ , so by the squeeze theorem $\dfrac{\sin x}{x}\to1$ .

✓Final answer
$\displaystyle\lim_{x\to 0}\dfrac{\sin x}{x}=1$
CBSE 2025Set ANNUAL1 markMCQ
$\lim_{x \to 0} \dfrac{\sin x}{x} =$
(a) 0
(b) 1
(c) -1
(d) $1/2$
›Reveal solutionSolution

$\lim_{x\to0}\dfrac{\sin x}{x}=1$ .

This is one of the standard trigonometric limits proved geometrically (comparing areas of triangles and a sector in a unit circle) in the NCERT text. It is used repeatedly to differentiate $\sin x$ , $\cos x$ , $\tan x$ , etc.

$\lim_{x\to0}\dfrac{\sin x}{x}=1$ .

✓Final answer
$(b) 1.$
CBSE 2025Set ANNUAL1 markMCQ
$\displaystyle\lim_{x \to 0}\left(\dfrac{\sin ax}{\sin bx}\right)$
(a) $\dfrac{b}{a}$
(b) $\dfrac{a}{b}$
(c) $\dfrac{-a}{b}$
(d) $\dfrac{-b}{a}$
›Reveal solutionSolution

As $x\to0$ , $\sin(ax)\approx ax$ and $\sin(bx)\approx bx$ , so the ratio tends to $a/b$ .

$\lim_{x\to0}\frac{\sin ax}{\sin bx} = \lim_{x\to0}\frac{\dfrac{\sin ax}{ax}\cdot ax}{\dfrac{\sin bx}{bx}\cdot bx} = \lim_{x\to0}\frac{\dfrac{\sin ax}{ax}}{\dfrac{\sin bx}{bx}}\cdot\frac{a}{b}$

Since $\lim_{x\to0}\dfrac{\sin ax}{ax} = 1$ and $\lim_{x\to0}\dfrac{\sin bx}{bx}=1$ (standard limit $\lim_{\theta\to0}\frac{\sin\theta}{\theta}=1$ , applied with $\theta=ax$ and $\theta=bx$ respectively):

$\lim_{x\to0}\frac{\sin ax}{\sin bx} = 1 \times \frac{a}{b} = \frac{a}{b}$

✓Final answer
$\dfrac{a}{b}$
CBSE 2025Set ANNUAL1 markMCQ
$\displaystyle\lim_{x \to 0}\left(\dfrac{3^{2x} - 1}{2^{3x} - 1}\right)$
(a) $\dfrac{\log 9}{\log 8}$
(b) $\dfrac{\log 8}{\log 9}$
(c) $\dfrac{\log 2}{\log 3}$
(d) $\dfrac{\log 3}{\log 2}$
›Reveal solutionSolution

Rewriting $3^{2x}=9^x$ and $2^{3x}=8^x$ and applying the standard limit $\lim_{x\to0}\frac{k^x-1}{x}=\log k$ to numerator and denominator gives $\dfrac{\log9}{\log8}$ .

$\lim_{x\to0}\frac{3^{2x}-1}{2^{3x}-1} = \lim_{x\to0}\frac{9^x-1}{8^x-1}$

Divide numerator and denominator by $x$ :

$= \lim_{x\to0}\frac{\dfrac{9^x-1}{x}}{\dfrac{8^x-1}{x}}$

Using the standard limit $\lim_{x\to0}\dfrac{a^x-1}{x} = \log a$ (natural log, or any consistent base):

$= \frac{\log 9}{\log 8}$

✓Final answer
$\dfrac{\log 9}{\log 8}$
CBSE 2025Set ANNUAL1 mark
$\displaystyle\lim_{x \to 0} \dfrac{\tan x}{x}$
›Reveal solutionSolution

The standard limit $\displaystyle\lim_{x \to 0}\dfrac{\tan x}{x} = 1$ follows from $\displaystyle\lim_{x\to 0}\dfrac{\sin x}{x}=1$ and $\displaystyle\lim_{x\to 0}\cos x = 1$ .

$\dfrac{\tan x}{x} = \dfrac{\sin x}{x} \cdot \dfrac{1}{\cos x}$ .

As $x \to 0$ : $\displaystyle\lim_{x\to 0}\dfrac{\sin x}{x} = 1$ (standard result) and $\displaystyle\lim_{x\to 0}\dfrac{1}{\cos x} = \dfrac{1}{1} = 1$ .

So $\displaystyle\lim_{x\to 0}\dfrac{\tan x}{x} = 1 \times 1 = 1$ .

✓Final answer
$\displaystyle\lim_{x \to 0} \dfrac{\tan x}{x} = 1$
CBSE 2024Set ANNUAL1 markMCQ
$\displaystyle\lim_{x \to \pi} \dfrac{\sin(\pi-x)}{\pi(\pi-x)} =$
(a) $\dfrac{1}{\pi}$
(b) $\dfrac{1}{\pi^2}$
(c) 1
(d) None of these
›Reveal solutionSolution

Substitute $t = \pi - x$ to turn this into the standard limit $\lim_{t\to0} \dfrac{\sin t}{t} = 1$ .

Let $t = \pi - x$ . As $x \to \pi$ , $t \to 0$ . Rewrite the limit:

$\lim_{x\to\pi} \frac{\sin(\pi-x)}{\pi(\pi-x)} = \lim_{t\to0} \frac{\sin t}{\pi t} = \frac{1}{\pi}\lim_{t\to0}\frac{\sin t}{t}$

Using the standard trigonometric limit $\displaystyle\lim_{t\to0}\frac{\sin t}{t} = 1$ :

$= \frac{1}{\pi} \times 1 = \frac{1}{\pi}$

✓Final answer
$\dfrac{1}{\pi}$
CBSE 2024Set ANNUAL1 mark
$Q. Find: lim(x\to0) (a^x - 1) / x = ?$
›Reveal solutionSolution

This is a standard limit result: $\lim_{x\to0}\frac{a^x-1}{x}=\ln a$ (natural log of $a$ ).

Write $a^x = e^{x\ln a}$ . Then

$\dfrac{a^x-1}{x} = \dfrac{e^{x\ln a}-1}{x} = \ln a\cdot\dfrac{e^{x\ln a}-1}{x\ln a}$ .

As $x\to 0$ , let $t=x\ln a\to 0$ , and using the standard limit $\lim_{t\to0}\frac{e^t-1}{t}=1$ :

$\lim_{x\to0}\dfrac{a^x-1}{x} = \ln a \times 1 = \ln a$ .

✓Final answer
$\displaystyle\lim_{x\to0}\frac{a^x-1}{x} = \ln a$
CBSE 2024Set ANNUAL1 markMCQ
$\displaystyle\lim_{x\to 0}\dfrac{a^x-b^x}{x}=$
(a) $\log\left(\dfrac{b}{a}\right)$
(b) $\log ab$
(c) $\dfrac{a}{b}$
(d) $\log\left(\dfrac{a}{b}\right)$
›Reveal solutionSolution

The limit equals $\log\left(\dfrac{a}{b}\right)$ .

Write the limit as a difference of two standard limits:

$\lim_{x\to0}\dfrac{a^x-b^x}{x}=\lim_{x\to0}\dfrac{a^x-1}{x}-\lim_{x\to0}\dfrac{b^x-1}{x}=\log a-\log b=\log\left(\dfrac{a}{b}\right),$

using the standard result $\displaystyle\lim_{x\to0}\dfrac{a^x-1}{x}=\log a$ .

✓Final answer
$\displaystyle\lim_{x\to0}\dfrac{a^x-b^x}{x}=\log\left(\dfrac{a}{b}\right)$
CBSE 2024Set ANNUAL1 markMCQ
$\displaystyle\lim_{x\to 0}\dfrac{e^x-1}{x} = \ldots\ldots$
(a) $0$
(b) $1$
(c) $\infty$
(d) $e^x$ .
›Reveal solutionSolution

$\displaystyle\lim_{x\to0}\dfrac{e^x-1}{x}=1$ .

Using the series expansion $e^x = 1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\cdots$ , we get

$e^x-1 = x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\cdots,$

so

$\dfrac{e^x-1}{x}=1+\dfrac{x}{2!}+\dfrac{x^2}{3!}+\cdots.$

As $x\to0$ , every term after the $1$ vanishes, giving the limit $1$ . This is a standard result used repeatedly in calculus.

✓Final answer
$1$
CBSE 2024Set ANNUAL1 markMCQ
$\displaystyle\lim_{x\to 0}\dfrac{\tan x}{x} = \ldots\ldots$
(a) $x$
(b) $1$
(c) $0$
(d) $\infty$ .
›Reveal solutionSolution

$\displaystyle\lim_{x\to0}\dfrac{\tan x}{x}=1$ .

Write $\tan x = \dfrac{\sin x}{\cos x}$ , so

$\dfrac{\tan x}{x}=\dfrac{\sin x}{x}\cdot\dfrac{1}{\cos x}.$

Using the standard limit $\displaystyle\lim_{x\to0}\dfrac{\sin x}{x}=1$ and $\displaystyle\lim_{x\to0}\cos x=1$ :

$\lim_{x\to0}\dfrac{\tan x}{x}=1\times\dfrac{1}{1}=1.$

✓Final answer
$1$

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