MHT-CET Maths · Limits

Exponential, Logarithmic and 1^∞ Limits

Three standard limits — (aˣ − 1)/x → log a, log(1 + x)/x → 1 and (1 + x)^(1/x) → e — plus the algebra that reduces a stem to them.

All Limits notes MHT-CET Maths strategy

Why this matters

21 PYQs at 71% HARD, the hardest page in the chapter by rate and still the most productive for marks per formula: in the 2021–2025 papers, five of the eleven were one factorisation or one substitution away from (aˣ − 1)/x. The 1 to the power infinity form appears both here and, more often, wearing a continuity costume on the next page — learn e to the power of the limit of (f − 1)g once and it answers both. In the CET bank and textbook log means the natural logarithm; a base is written only when it is not e.

Concept 1 of 6: The Exponential and Logarithmic Standard Limits

Near

0

a^x - 1

grows like

x\log a

— the slope of

a^x

at the origin — and

\log(1 + x)

grows like

x

. Each standard limit is one of these slopes.

Definition

$\lim_{x\to 0}\dfrac{a^x - 1}{x} = \log a$ ; in particular $\lim_{x\to 0}\dfrac{e^x - 1}{x} = 1$ .
$\lim_{x\to 0}\dfrac{\log(1 + x)}{x} = 1$ , and scaled: $\lim_{x\to 0}\dfrac{\log(1 + kx)}{x} = k$ , $\lim_{x\to 0}\dfrac{\log(1 + x^2/3)}{x^2} = \dfrac{1}{3}$ .
Replacement rule, exactly as for $\sin$ : inside a product or quotient, $a^{\square} - 1 \to \square\log a$ , $e^{\square} - 1 \to \square$ , $\log(1 + \square) \to \square$ whenever $\square \to 0$ .
Then count powers of $x$ on each floor; matched powers give a finite constant, mismatched give $0$ or $\infty$ .
$\log a - \log b = \log\dfrac{a}{b}$ and $\log a^2 = 2\log a$ are how the option list is written — reduce your answer to a single log.

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 $e$ )

Worked example

Evaluate

\lim_{x\to 0}\dfrac{3^x - 1}{2^x - 1}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 10th May Shift 2 · Q138Moderate

Example 1 · Limits · Exponential, Logarithmic and 1^∞ Limits

\lim_{x\to 0}\frac{9^x - 4^x}{x(9^x + 4^x)} =

(aˣ − 1)/x is log a, never a

The limit is the slope of

a^x

0

, which is

\log a

. An option reading

\dfrac{3}{2}

where

\log\dfrac{3}{2}

is correct is the standard distractor.

Drill 1 more on the exponential and logarithmic standard limits

Concept 2 of 6: Factorising aˣ − bˣ − cˣ + 1 into (bˣ − 1)(cˣ − 1)

63^x = 9^x\cdot 7^x

. So

63^x - 9^x - 7^x + 1

9^x\cdot7^x - 9^x - 7^x + 1

, which factors by grouping into

(9^x - 1)(7^x - 1)

— two standard limits multiplied together.

Definition

Whenever the first base is the product of the other two: $(bc)^x - b^x - c^x + 1 = (b^x - 1)(c^x - 1)$ .
The product is second order in $x$ : $(b^x - 1)(c^x - 1) \sim x^2\log b\log c$ . So it pairs with an $x^2$ , an $x\sin x$ , or a $1 - \cos x$ below.
Variants: $a^x - b^x = (a^x - 1) - (b^x - 1) \sim x(\log a - \log b) = x\log\dfrac{a}{b}$ ; and $10^x + 7^x - 14^x - 5^x = -(2^x - 1)(7^x - 5^x)$ ... check the grouping by expanding it back.
$1^x$ in a stem is just $1$ : $8^x - 4^x - 2^x + 1^x = (4^x - 1)(2^x - 1)$ .

Grouping factorisation

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

Worked example

Evaluate

\lim_{x\to 0}\dfrac{6^x - 3^x - 2^x + 1}{x^2}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 25 April Shift II · Q104Hard

Example 2 · Limits · Exponential, Logarithmic and 1^∞ Limits

\lim_{x\rightarrow 0} \frac{63^{x}-9^{x}-7^{x}+ 1}{\sqrt{2}-\sqrt{1 + \cos x}}=

...... .

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

1 - \cos x

, not with

x

Concept 3 of 6: Substitute t = aˣ When the Exponents Are Mixed

5^x

5^{3 - x}

and

5^{x/2}

are all powers of one thing. Call

5^{x/2} = t

and the whole limit becomes a rational function of

t

— a factor-and-cancel problem from the algebraic page.

Definition

Choose $t$ as the smallest power present so that every other exponential is an integer power of $t$ : with $5^{x/2} = t$ , $5^x = t^2$ and $5^{3 - x} = \dfrac{125}{t^2}$ .
Translate the point: $x \to 2$ becomes $t \to 5$ .
Clear the fractions in $t$ , factor the polynomial (it vanishes at the new point), cancel, substitute — or use the derivative-in-disguise reading on the $t$ -expression.
Same idea for $4^x$ with $2^x$ : put $2^x = t$ , so $4^x = t^2$ and $2^{x + 1} = 2t$ .

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}

Worked example

Evaluate

\lim_{x\to 1}\dfrac{4^x - 2^{x + 1}}{2^x - 2}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 11th May Shift 1 · Q105Hard

Example 3 · Limits · Exponential, Logarithmic and 1^∞ Limits

\lim_{x \to 2} \frac{5^x + 5^{3-x} - 30}{5^{3-x} - 5^{x/2}} =

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.

Drill 1 more on substitute t = aˣ when the exponents are mixed

Concept 4 of 6: Composite Forms: (eᵘ − 1)/u with u → 0, and Mixed Series Terms

e^{\tan x} - e^{x}

is not a standard form until you factor out

e^{x}

: then it is

e^{x}(e^{u} - 1)

with

u = \tan x - x \to 0

, and

\dfrac{e^{u} - 1}{u} \to 1

even though

u

is a complicated function.

Definition

$e^{a} - e^{b} = e^{b}(e^{a - b} - 1)$ , and $\dfrac{e^{a - b} - 1}{a - b} \to 1$ whenever $a - b \to 0$ . This kills any $\dfrac{e^{f(x)} - e^{g(x)}}{f(x) - g(x)}$ instantly.
The same rule for any inner function: $\dfrac{e^{x^2} - 1}{x^2} \to 1$ , $\dfrac{e^{\sin x} - 1}{\sin x} \to 1$ , $\dfrac{e^{5\sqrt{x}} - 1}{5\sqrt{x}} \to 1$ .
Mixed numerators like $e^{x^2} - \cos 3x$ : add and subtract $1$ to split into $(e^{x^2} - 1) + (1 - \cos 3x)$ , each a known order — $x^2$ and $\dfrac{9x^2}{2}$ .
Then the denominator's pieces are replaced the same way: $\sin x\log(1 + 2x) \sim x\cdot 2x = 2x^2$ .

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

Worked example

Evaluate

\lim_{x\to 0}\dfrac{e^{\sin x} - e^{x}}{\sin x - x}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 20 April Shift I · Q124Moderate

Example 4 · Limits · Exponential, Logarithmic and 1^∞ Limits

\lim_{x\rightarrow 0} \frac{e^{\tan x}-e^{x}}{\tan x-x}=

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.

Drill 2 more on composite forms: (eᵘ − 1)/u with u → 0, and mixed series terms

Concept 5 of 6: The 1^∞ Form: lim f^g = e^{lim (f − 1)g}

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

is 'almost 1' raised to a huge power, and the two effects balance to give

e

. Every

1^\infty

limit is that balance in disguise: the answer is

e

raised to how fast the base leaves

1

times how fast the exponent grows.

Definition

Base limits: $\lim_{x\to 0}(1 + x)^{1/x} = e$ and $\lim_{x\to\infty}\left(1 + \dfrac{k}{x}\right)^{x} = e^{k}$ .
General rule: if $f \to 1$ and $g \to \infty$ , then $\lim f^{g} = e^{\lim (f - 1)\,g}$ . Compute the exponent $(f - 1)g$ as an ordinary limit.
Recognise the form first: $\dfrac{x + 8}{x + 1} \to 1$ with exponent $x + 5 \to \infty$ ; $\log_3 3x = 1 + \log_3 x \to 1$ with exponent $\log_x 8 \to \infty$ ; $(1 + \tan x)^{\csc x}$ at $0$ .
Quotient shortcut: $\dfrac{(1 + 8/x)^{x}}{(1 + 1/x)^{x}} \to \dfrac{e^{8}}{e^{1}} = e^{7}$ .
If the base does not tend to $1$ the form is not $1^\infty$ : $\left(\frac45\right)^{\text{something} \to 0}$ is simply $1$ , and $\left(\frac45\right)^{\text{something}\to\infty}$ is $0$ .

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}

Worked example

Evaluate

\lim_{x\to\infty}\left(\dfrac{x + 3}{x - 1}\right)^{x}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 22 April Shift I · Q134Moderate

Example 5 · Limits · Exponential, Logarithmic and 1^∞ Limits

\lim_{x\rightarrow \infty} \left( \frac{x+ 8}{x+ 1} \right)^{x+ 5}= \ldots

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.

Drill 1 more on the 1^∞ form: lim f^g = e^{lim (f − 1)g}

Concept 6 of 6: 0⁰ and ∞⁰ Forms — Take Logarithms

x^{x}

x \to 0^+

pits a base going to

0

against an exponent going to

0

. Taking logs turns the power into a product,

x\log x

, and that product tends to

0

— the log loses to the power of

x

Definition

For $l = \lim f^{g}$ with $f \to 0^+, g \to 0$ (form $0^0$ ) or $f \to \infty, g \to 0$ (form $\infty^0$ ): compute $\log l = \lim g\log f$ , then $l = e^{\log l}$ .
Key fact: $\lim_{x\to 0^+} x\log x = 0$ — any positive power of $x$ beats $\log x$ . Hence $x^{x} \to 1$ , $x^{\sin x} \to 1$ , $x^{2x} \to 1$ .
Not every power form is indeterminate: $(\sin x)^{1/x}$ as $x \to 0^+$ has base $\to 0$ and exponent $\to +\infty$ , so it is simply $0$ — a tiny number raised to a huge power.
Sums of such terms are evaluated term by term once each is settled: $(\sin x)^{1/x} + \dfrac{1}{x^{\sin x}} \to 0 + 1 = 1$ .

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

Worked example

Evaluate

\lim_{x\to 0^+} x^{x}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 16th May Shift 1 · Q116Hard

Example 6 · Limits · Exponential, Logarithmic and 1^∞ Limits

The value of

\lim_{x \to 0^+} \left[(\sin x)^{1/x} + \frac{1}{x^{\sin x}}\right]

where

x > 0

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.

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)

The Exponential and Logarithmic Standard Limits
Exponential and logarithmic standard limits
$\lim_{x\to 0}\frac{a^x - 1}{x} = \log a \qquad \lim_{x\to 0}\frac{e^x - 1}{x} = 1 \qquad \lim_{x\to 0}\frac{\log(1 + x)}{x} = 1 \qquad \lim_{x\to 0}\frac{\log(1 + kx)}{x} = k$
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
$l = \lim f^{g} \ \Rightarrow\ \log l = \lim g\log f \qquad \lim_{x\to 0^+} x\log x = 0 \ \Rightarrow\ \lim_{x\to 0^+} x^{x} = 1$

Watch out for (6)

(aˣ − 1)/x is log a, never a
→ The Exponential and Logarithmic Standard Limits
Sending each term to its own limit
→ Factorising aˣ − bˣ − cˣ + 1 into (bˣ − 1)(cˣ − 1)
Choosing t too large
→ Substitute t = aˣ When the Exponents Are Mixed
Using only first order on eˣ² − cos x
→ Composite Forms: (eᵘ − 1)/u with u → 0, and Mixed Series Terms
1^∞ is indeterminate — never answer 1 by substitution
→ The 1^∞ Form: lim f^g = e^{lim (f − 1)g}
Treating (sin x)^(1/x) as a 0⁰ form
→ 0⁰ and ∞⁰ Forms — Take Logarithms

Test yourself on Limits

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