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Advanced Calculus Reading Time: 15 min 4 Key Expansions

Expansion Series Method: Using Taylor & Maclaurin Series

Master sin x, cos x, e^x, ln(1+x) expansions to solve complex limits that stump most JEE students.

4
Key Series
90%
JEE Relevance
3-5
Terms Needed
5min
Avg. Solve Time

Why Series Expansion is a Game-Changer for JEE Limits

When direct substitution gives $\frac{0}{0}$ or other indeterminate forms, series expansion provides the most powerful method to evaluate limits. This technique appears in 2-3 questions per JEE paper and can save you 5-10 minutes on complex problems.

Taylor & Maclaurin Series Definition

Taylor Series about $x = a$:

$$ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots $$

Maclaurin Series (special case when $a = 0$):

$$ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots $$

🎯 Series Navigation

sin x cos x e^x ln(1+x)
Maclaurin Series Converges for all x

Sine Function: $\sin x$

Series Expansion:

$$ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} - \cdots $$

General term: $(-1)^n \frac{x^{2n+1}}{(2n+1)!}$ for $n = 0, 1, 2, \ldots$

📝 JEE Application Example

Problem: Evaluate $\lim_{x \to 0} \frac{\sin x - x}{x^3}$

Step 1: Expand sin x

Using Maclaurin series:

$$ \sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots $$
Step 2: Substitute in limit
$$ \frac{\sin x - x}{x^3} = \frac{(x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots) - x}{x^3} = \frac{-\frac{x^3}{6} + \frac{x^5}{120} - \cdots}{x^3} $$
Step 3: Simplify and take limit
$$ = -\frac{1}{6} + \frac{x^2}{120} - \cdots \xrightarrow[x \to 0]{} -\frac{1}{6} $$

Final Answer: $-\frac{1}{6}$

💡 JEE Strategy for sin x

  • For limits as $x \to 0$, use first 2-3 terms: $\sin x \approx x - \frac{x^3}{6}$
  • Remember: $\sin x \approx x$ for very small x (first approximation)
  • For $\sin(ax)$ expansion, replace x with ax in the series
  • Watch for cancellation with other series expansions
Maclaurin Series Converges for all x

Cosine Function: $\cos x$

Series Expansion:

$$ \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \cdots $$

General term: $(-1)^n \frac{x^{2n}}{(2n)!}$ for $n = 0, 1, 2, \ldots$

📝 JEE Application Example

Problem: Evaluate $\lim_{x \to 0} \frac{1 - \cos x}{x^2}$

Step 1: Expand cos x

Using Maclaurin series:

$$ \cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720} + \cdots $$
Step 2: Substitute in limit
$$ \frac{1 - \cos x}{x^2} = \frac{1 - (1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots)}{x^2} = \frac{\frac{x^2}{2} - \frac{x^4}{24} + \cdots}{x^2} $$
Step 3: Simplify and take limit
$$ = \frac{1}{2} - \frac{x^2}{24} + \cdots \xrightarrow[x \to 0]{} \frac{1}{2} $$

Final Answer: $\frac{1}{2}$

💡 JEE Strategy for cos x

  • For limits as $x \to 0$, use: $\cos x \approx 1 - \frac{x^2}{2}$
  • Remember: $\cos x \approx 1$ for very small x (crude approximation)
  • For $\cos(ax)$ expansion, replace x with ax in the series
  • Often used with sin x expansion in the same problem
Maclaurin Series Converges for all x

Exponential Function: $e^x$

Series Expansion:

$$ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \cdots $$

General term: $\frac{x^n}{n!}$ for $n = 0, 1, 2, \ldots$

📝 JEE Application Example

Problem: Evaluate $\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}$

Step 1: Expand e^x

Using Maclaurin series:

$$ e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \cdots $$
Step 2: Substitute in limit
$$ \frac{e^x - 1 - x}{x^2} = \frac{(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots) - 1 - x}{x^2} = \frac{\frac{x^2}{2} + \frac{x^3}{6} + \cdots}{x^2} $$
Step 3: Simplify and take limit
$$ = \frac{1}{2} + \frac{x}{6} + \cdots \xrightarrow[x \to 0]{} \frac{1}{2} $$

Final Answer: $\frac{1}{2}$

💡 JEE Strategy for e^x

  • For limits as $x \to 0$, use: $e^x \approx 1 + x + \frac{x^2}{2}$
  • Remember: $e^x \approx 1 + x$ for very small x (linear approximation)
  • For $e^{ax}$ expansion, replace x with ax in the series
  • Very useful in limits involving growth and decay functions
Maclaurin Series Converges for |x| < 1

Logarithm Function: $\ln(1+x)$

Series Expansion:

$$ \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \cdots $$

General term: $(-1)^{n+1} \frac{x^n}{n}$ for $n = 1, 2, 3, \ldots$

⚠️ Convergence Warning

This series converges only for $-1 < x \leq 1$. For limits as $x \to 0$, this is always satisfied.

📝 JEE Application Example

Problem: Evaluate $\lim_{x \to 0} \frac{\ln(1+x) - x}{x^2}$

Step 1: Expand ln(1+x)

Using Maclaurin series:

$$ \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots $$
Step 2: Substitute in limit
$$ \frac{\ln(1+x) - x}{x^2} = \frac{(x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots) - x}{x^2} = \frac{-\frac{x^2}{2} + \frac{x^3}{3} - \cdots}{x^2} $$
Step 3: Simplify and take limit
$$ = -\frac{1}{2} + \frac{x}{3} - \cdots \xrightarrow[x \to 0]{} -\frac{1}{2} $$

Final Answer: $-\frac{1}{2}$

💡 JEE Strategy for ln(1+x)

  • For limits as $x \to 0$, use: $\ln(1+x) \approx x - \frac{x^2}{2}$
  • Remember: $\ln(1+x) \approx x$ for very small x (first approximation)
  • Watch for domain restrictions in non-limit problems
  • Often appears in limits with other transcendental functions

🚀 Advanced JEE Applications

Combining Multiple Expansions

Many JEE problems require using 2+ expansions simultaneously:

Example: $\lim_{x \to 0} \frac{e^x \sin x - x(1+x)}{x^3}$

Solution: Expand both $e^x$ and $\sin x$, multiply series, then simplify

Taylor Series at Non-zero Points

For limits as $x \to a$ where $a \neq 0$:

Example: $\lim_{x \to 1} \frac{\ln x - (x-1)}{(x-1)^2}$

Solution: Use Taylor series of ln x about x=1

📋 Quick Reference: Essential Expansions

Function Expansion (up to x³) Convergence JEE Usage
sin x $x - \frac{x^3}{6}$ All x Very High
cos x $1 - \frac{x^2}{2}$ All x Very High
e^x $1 + x + \frac{x^2}{2} + \frac{x^3}{6}$ All x High
ln(1+x) $x - \frac{x^2}{2} + \frac{x^3}{3}$ |x| < 1 Medium

🎯 Practice Problems

Test your understanding with these JEE-style problems:

1. $\lim_{x \to 0} \frac{e^x \cos x - 1 - x}{x^2}$

Hint: Expand both e^x and cos x, then multiply series

2. $\lim_{x \to 0} \frac{\sin x - x \cos x}{x^3}$

Hint: Use sin x and cos x expansions, watch for cancellation

3. $\lim_{x \to 0} \frac{\ln(1 + \sin x)}{x}$

Hint: Expand sin x first, then use ln(1+u) expansion

💡 Mastery Tips for JEE

Exam Strategy

  • Memorize the first 3 terms of each expansion
  • Practice identifying when to use series expansion vs other methods
  • Always check if direct substitution gives indeterminate form
  • Use expansion when L'Hôpital's rule becomes too messy

Common Pitfalls

  • Using too few terms (missing the answer)
  • Using ln(1+x) expansion outside its convergence radius
  • Forgetting to distribute negative signs
  • Mishandling factorial calculations

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