Evaluate $\displaystyle\lim_{x\to 0}\frac{\log_e (1 + 4x)}{x}$.

Evaluate limx→ 0(loge (1 + 4x))/(x).

Worked Examples · Example 7

\displaystyle\lim_{x\to 0}\frac{\log_e (1 + 4x)}{x}

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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. …

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At $x=0$ : $\tfrac{\log_e 1}{0}=\tfrac{0}{0}$ . Use $\lim_{u\to0}\dfrac{\log_e(1+u)}{u}=1$ with $u=4x$ :

$\frac{\log_e(1+4x)}{x}=4\cdot\frac{\log_e(1+4x)}{4x}\xrightarrow{\ x\to0\ }4\cdot1=4.$ …

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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 $2x$ .

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

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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

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

…

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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 $0<x<\tfrac{\pi}{2}$ .

…

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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 differenti …

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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 $a/b$ .

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

…

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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 $x$ :

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

…

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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 $\displaystyle\lim_{x\to 0}\cos x = 1$ .

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

…

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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:

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

…

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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

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

…

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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:

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

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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

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

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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

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

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Q.Evaluate lim⁡x→0log⁡e(1+4x)x\displaystyle\lim_{x\to 0}\frac{\log_e (1 + 4x)}{x}x→0lim​xloge​(1+4x)​.

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Q.Evaluate $\displaystyle\lim_{x\to 0}\frac{\log_e (1 + 4x)}{x}$ .

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