Limits

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Definition: $\lim_{x \to a} f(x) = L$ means that as $x$ gets arbitrarily close to $a$ (but not equal to $a$), $f(x)$ gets arbitrarily close to $L$.

Standard Limits:

1. $\lim_{x \to 0} \frac{\sin x}{x} = 1$ (where $x$ is in radians)
2. $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$
3. $\lim_{x \to 0} \frac{\tan x}{x} = 1$
4. $\lim_{x \to 0} \frac{e^x - 1}{x} = 1$
5. $\lim_{x \to 0} \frac{\ln(1+x)}{x} = 1$
6. $\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x = e$

Indeterminate Forms: $frac{0}{0}, \frac{\infty}{\infty}, 0 \times \infty, \infty - \infty, 0^0, \infty^0, 1^\infty$

⚡ JEE Tip: For $\frac{0}{0}$ forms, factor and cancel, or use series expansion. For $\frac{\infty}{\infty}$, divide numerator and denominator by highest power of $x$.

⚡ Common Mistake: $\frac{\sin x}{x}$ limit only equals 1 when $x \to 0$ and $x$ is in radians. Always convert degrees to radians first.

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🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding.

Methods for Evaluating Limits:

1. Factorisation Method: For $\frac{0}{0}$ forms, if $f(a) = 0$ and $g(a) = 0$, factor $(x-a)$ from both numerator and denominator.

2. Rationalisation: Use when expressions involve square roots.

3. L’Hôpital’s Rule: If $\lim_{x \to a} \frac{f(x)}{g(x)}$ gives $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then: $$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f’(x)}{g’(x)}$$ Can be applied repeatedly if needed (but check conditions each time).

4. Series Expansion:

• $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$
• $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$
• $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$
• $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$
• $\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + \cdots$

Special Limit Forms:

Sandwich Theorem: If $f(x) \leq g(x) \leq h(x)$ near $a$ and $\lim f(x) = \lim h(x) = L$, then $\lim g(x) = L$.

Exponential Limit: $\lim_{x \to 0} (1 + x)^{1/x} = e$.

Worked Examples:

Example 1: Evaluate $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$. Factor: $\frac{(x+2)(x-2)}{x-2} = x + 2$. At $x = 2$: limit $= 4$.

Example 2: Evaluate $\lim_{x \to 0} \frac{\sin 5x}{x}$. Using $\lim_{x \to 0} \frac{\sin x}{x} = 1$: $\frac{\sin 5x}{x} = \frac{\sin 5x}{5x} \cdot 5$. As $x \to 0$, $\frac{\sin 5x}{5x} \to 1$. So limit $= 5$.

Example 3 (JEE 2021): Evaluate $\lim_{x \to 0} \frac{\cos x - 1 + \frac{x^2}{2}}{x^4}$. Using Maclaurin series: $\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots$. So $\cos x - 1 + \frac{x^2}{2} = \frac{x^4}{24} + O(x^6)$. Dividing by $x^4$: limit $= \frac{1}{24}$.

Example 4: Evaluate $\lim_{x \to \infty} \frac{x^2 + 3x}{2x^2 + 5}$. Divide numerator and denominator by $x^2$: $\frac{1 + \frac{3}{x}}{2 + \frac{5}{x^2}} \to \frac{1}{2}$ as $x \to \infty$.

Example 5: Evaluate $\lim_{x \to 0} \frac{e^x - e^{-x}}{\sin x}$. Using series: $e^x - e^{-x} = (1+x+\frac{x^2}{2}+\cdots) - (1-x+\frac{x^2}{2}-\cdots) = 2x + O(x^3)$. $\sin x = x - \frac{x^3}{6} + O(x^5)$. So $\frac{2x + O(x^3)}{x + O(x^3)} \to 2$ as $x \to 0$.

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🔴 Extended — Deep Study (3mo+)

Comprehensive theory for serious JEE Advanced preparation.

Extended Algebraic Manipulations:

For limits of form $1^\infty$: $$\lim_{x \to a} [f(x)]^{g(x)} = e^{\lim_{x \to a} (f(x) - 1) \cdot g(x)}$$

For forms like $\infty^0$ or $0^0$, take logarithm: Let $L = \lim f(x)^{g(x)}$, then $\ln L = \lim g(x) \cdot \ln f(x)$.

Standard Results for Reference:

• $\lim_{x \to 0} \frac{a^x - 1}{x} = \ln a$ for $a > 0$
• $\lim_{x \to 0} \frac{(1+x)^n - 1}{x} = n$
• $\lim_{x \to 0} \frac{\tan x - \sin x}{x^3} = \frac{1}{2}$ (derived from series)
• $\lim_{x \to 0} \frac{\sin x - x}{x^3} = -\frac{1}{6}$
• $\lim_{x \to 0} \frac{\cos x - 1 + \frac{x^2}{2}}{x^4} = \frac{1}{24}$

Nth Derivative Using Limits:

$f^{(n)}(a) = \lim_{h \to 0} \frac{f^{(n-1)}(a+h) - f^{(n-1)}(a)}{h}$, applied recursively.

Continuity and Differentiability Connection:

If $\lim_{x \to a^+} f(x) = \lim_{x \to a^-} f(x) = f(a)$, then $f$ is continuous at $a$. If $f$ is differentiable at $a$, it must be continuous at $a$ (but not conversely).

Advanced Limit Problems:

Problem 1 (JEE Advanced 2019): Evaluate $\lim_{n \to \infty} \frac{1 + 2 + 3 + \cdots + n}{n^2}$.

$\lim_{n \to \infty} \frac{n(n+1)}{2n^2} = \lim_{n \to \infty} \frac{n+1}{2n} = \frac{1}{2}$.

Problem 2: Evaluate $\lim_{x \to 0} \frac{\sin(\pi \sin x)}{x}$.

As $x \to 0$, $\sin x \approx x$, so $\sin(\pi \sin x) \approx \sin(\pi x) \approx \pi x$. So limit $\approx \frac{\pi x}{x} = \pi$.

Problem 3: Evaluate $\lim_{x \to 0} \left(\frac{\sin x}{x}\right)^{1/x^2}$.

Let $L = \lim_{x \to 0} \left(\frac{\sin x}{x}\right)^{1/x^2}$. $\ln L = \lim_{x \to 0} \frac{\ln(\sin x) - \ln(x)}{x^2}$. Using series: $\sin x = x - \frac{x^3}{6} + \cdots$. $\ln(\sin x) = \ln x + \ln(1 - \frac{x^2}{6} + \cdots) = \ln x - \frac{x^2}{6} + O(x^4)$. So $\ln(\sin x) - \ln x = -\frac{x^2}{6} + O(x^4)$. Then $\ln L = \lim_{x \to 0} \frac{-x^2/6 + O(x^4)}{x^2} = -\frac{1}{6}$. So $L = e^{-1/6}$.

Problem 4 (L’Hôpital repeated): Evaluate $\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}$.

Direct substitution gives $\frac{0}{0}$. Apply L’Hôpital: Derivative numerator: $e^x - 1$; derivative denominator: $2x$. New limit: $\lim_{x \to 0} \frac{e^x - 1}{2x} = \frac{1}{2}$ (since $\lim_{x \to 0} \frac{e^x - 1}{x} = 1$).

JEE Advanced Patterns (2018–2024):

• Series expansion limits appear frequently (especially with $\sin$, $\cos$, $e^x$)
• $1^\infty$ form has appeared multiple times in recent papers
• Repeated L’Hôpital applications tested in 2020, 2022
• Limits involving trigonometric functions and logarithms in combination are common
• Sandwich theorem used in 2021 for non-polynomial bounds

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