The $1^\infty$ Form: The Complete Shortcut

Introduction

The $1^\infty$ indeterminate form appears in 25–30 JEE Main questions over the past six years, making it one of the most reliably tested limit types. Despite looking intimidating, every $1^\infty$ problem can be solved with a single universal formula. Master this shortcut and you convert a potentially complex problem into a 60-second calculation.

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1. Fundamental Principle

Why is $1^\infty$ Indeterminate?

Consider $f(x)^{g(x)}$ where $f(x) \to 1$ and $g(x) \to \infty$. The result is not necessarily 1:

• $\left(1 + \frac{1}{n}\right)^n \to e \approx 2.718$
• $\left(1 + \frac{2}{n}\right)^n \to e^2$
• $\left(1 + \frac{1}{n^2}\right)^n \to 1$

The competition between "base approaching 1" and "exponent approaching infinity" determines the outcome.

The Universal Formula

If $\lim_{x \to a} f(x) = 1$ and $\lim_{x \to a} g(x) = \infty$, then:

$\boxed{\lim_{x \to a} f(x)^{g(x)} = e^{\displaystyle \lim_{x \to a} g(x)[f(x) - 1]}}$

Proof sketch: Write $f(x)^{g(x)} = \left[1 + (f(x)-1)\right]^{g(x)}$. Let $h = f(x) - 1 \to 0$. Then: $[1+h]^{g(x)} = \left[(1+h)^{1/h}\right]^{h \cdot g(x)} \to e^{\lim h \cdot g(x)}.$

The Three-Step Recipe

Step 1: Confirm the form is $1^\infty$ (base $\to 1$, exponent $\to \infty$).

Step 2: Compute $L = \lim_{x \to a} g(x) \cdot [f(x) - 1]$.

Step 3: The answer is $e^L$.

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2. Common Disguises

The $1^\infty$ form often appears in disguise. Watch for these patterns:

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3. Worked Examples

Example 1 (Classic Template)

Evaluate $\displaystyle \lim_{n \to \infty} \left(1 + \frac{3}{n}\right)^{2n}$.

Solution:

Base $\to 1$, exponent $\to \infty$. Apply the formula: $L = \lim_{n \to \infty} 2n \cdot \frac{3}{n} = 6.$ Answer: $e^6$.

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Example 2 (Trigonometric Base)

Evaluate $\displaystyle \lim_{x \to 0} (\cos x)^{1/x^2}$.

Solution:

Base $= \cos x \to 1$, exponent $= 1/x^2 \to \infty$. So: $L = \lim_{x \to 0} \frac{1}{x^2}(\cos x - 1) = \lim_{x \to 0} \frac{-2\sin^2(x/2)}{x^2} = \lim_{x \to 0} \frac{-2 \cdot x^2/4}{x^2} = -\frac{1}{2}.$

Answer: $e^{-1/2} = \dfrac{1}{\sqrt{e}}$.

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Example 3 (Rational Base, $x \to \infty$)

Evaluate $\displaystyle \lim_{x \to \infty} \left(\frac{x+3}{x-1}\right)^{x+1}$.

Solution:

Base $= \frac{x+3}{x-1} = 1 + \frac{4}{x-1} \to 1$. Exponent $= x + 1 \to \infty$. $L = \lim_{x \to \infty} (x+1) \cdot \frac{4}{x-1} = 4 \lim_{x \to \infty} \frac{x+1}{x-1} = 4.$

Answer: $e^4$.

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Example 4 (Log-based Exponent)

Evaluate $\displaystyle \lim_{x \to 0} (1 + \sin x)^{\cot x}$.

Solution:

Base $= 1 + \sin x \to 1$. Exponent $= \cot x = \frac{\cos x}{\sin x} \to \infty$. $L = \lim_{x \to 0} \cot x \cdot \sin x = \lim_{x \to 0} \cos x = 1.$

Answer: $e^1 = e$.

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Example 5 ($\sin x / x$ Base)

Evaluate $\displaystyle \lim_{x \to 0} \left(\frac{\sin x}{x}\right)^{1/x^2}$.

Solution:

Base $\to 1$, exponent $\to \infty$. $L = \lim_{x \to 0} \frac{1}{x^2} \left(\frac{\sin x}{x} - 1\right) = \lim_{x \to 0} \frac{\sin x - x}{x^3}.$ Using $\sin x = x - \frac{x^3}{6} + \cdots$: $L = \lim_{x \to 0} \frac{-x^3/6}{x^3} = -\frac{1}{6}.$

Answer: $e^{-1/6}$.

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Example 6 (Double Application)

Evaluate $\displaystyle \lim_{x \to 0} \left(\frac{\tan x}{x}\right)^{1/x^2}$.

Solution:

Base $\to 1$, exponent $\to \infty$. $L = \lim_{x \to 0} \frac{1}{x^2} \left(\frac{\tan x}{x} - 1\right) = \lim_{x \to 0} \frac{\tan x - x}{x^3}.$ Using $\tan x = x + \frac{x^3}{3} + \cdots$: $L = \frac{1}{3}.$

Answer: $e^{1/3}$.

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4. Previous Year JEE Problems

Problem 1 (JEE Main 2021)

$\displaystyle \lim_{n \to \infty} \left(\frac{n+1}{n-1}\right)^{n} =$ ?

Solution:

Base $= 1 + \frac{2}{n-1} \to 1$. Exponent $= n \to \infty$. $L = \lim_{n \to \infty} n \cdot \frac{2}{n-1} = \lim_{n \to \infty} \frac{2n}{n-1} = 2.$

Answer: $e^2$.

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Problem 2 (JEE Main 2023)

If $\displaystyle \lim_{x \to 0} \left(\frac{e^{2x} + e^x + 1}{3}\right)^{1/(2x)} = e^k$, find $k$.

Solution:

Check base: At $x = 0$, $\frac{1+1+1}{3} = 1$. Exponent $= \frac{1}{2x} \to \infty$. Confirmed $1^\infty$. $L = \lim_{x \to 0} \frac{1}{2x} \left(\frac{e^{2x}+e^x+1}{3} - 1\right) = \lim_{x \to 0} \frac{e^{2x}+e^x+1-3}{6x}.$ $= \lim_{x \to 0} \frac{(e^{2x}-1)+(e^x-1)}{6x} = \frac{2+1}{6} = \frac{1}{2}.$

So $k = \dfrac{1}{2}$.

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Problem 3 (JEE Main 2024)

$\displaystyle \lim_{x \to 0} \left(\frac{1 + \tan x}{1 + \sin x}\right)^{1/x^3} =$ ?

Solution:

Base $\to 1$. Exponent $\to \infty$. This is a $1^\infty$ form requiring series expansion. $L = \lim_{x \to 0} \frac{1}{x^3} \left(\frac{1+\tan x}{1+\sin x} - 1\right) = \lim_{x \to 0} \frac{1}{x^3} \cdot \frac{\tan x - \sin x}{1+\sin x}.$ Since $1 + \sin x \to 1$, we focus on: $L = \lim_{x \to 0} \frac{\tan x - \sin x}{x^3}.$ Using $\tan x - \sin x = \sin x\left(\frac{1}{\cos x} - 1\right) = \frac{\sin x(1-\cos x)}{\cos x}$.

Approximating for $x \to 0$: $\sin x \approx x$, $1-\cos x \approx x^2/2$, $\cos x \to 1$: $L = \lim_{x \to 0} \frac{x \cdot (x^2/2)}{x^3} = \frac{1}{2}.$

Answer: $e^{1/2} = \sqrt{e}$.

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5. Common $1^\infty$ Results to Memorize

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6. Tips for JEE Aspirants

1. One formula to rule them all: $e^{\lim g(x)[f(x)-1]}$. Memorize this and you can solve any $1^\infty$ problem.
2. Always confirm the form. Check that the base really tends to 1 and the exponent really tends to infinity.
3. When computing $f(x) - 1$, simplify the fraction to a single expression first, then evaluate the limit with $g(x)$.
4. Use Taylor expansions when $f(x) - 1$ produces another $0/0$ form. Usually 2 terms suffice.
5. Watch for the sign of the exponent $L$—a negative value gives a result less than 1 (like $1/\sqrt{e}$).
6. Don't take logs unnecessarily. The direct formula avoids the log-antilog two-step.

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7. Quick-Reference Summary

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The $1^\infty$ form is a JEE favourite precisely because the formula is simple but students forget it under pressure. Drill 10–15 problems and the pattern becomes automatic.