The Hook: The Mystery of sin(x)/x
Imagine Virat Kohli hitting a six in Jawan (2023) — the ball follows a smooth arc. To find the exact direction at the highest point, we need calculus. But calculating the derivative of $\sin x$ requires knowing:
$$\lim_{x \to 0} \frac{\sin x}{x} = ?$$This limit can’t be solved by substitution (gives $\frac{0}{0}$), but it’s so fundamental that all of trigonometric calculus depends on it.
Today, we master the standard limits — the special formulas that unlock complex limit problems in seconds!
The Core Concept
What are Standard Limits?
Standard limits are pre-proven limit formulas that appear repeatedly in calculus. Instead of deriving them each time, we memorize and apply them directly.
Think of them as the “formulas” of limits — just like $a^2 - b^2 = (a+b)(a-b)$ in algebra.
Category 1: Trigonometric Limits
The Fundamental Trigonometric Limit
$$\boxed{\lim_{x \to 0} \frac{\sin x}{x} = 1}$$Where: $x$ is in radians, not degrees!
Interactive Demo: Visualize Standard Limits
Explore how standard limit formulas behave as x approaches specific values.
Proof Sketch: Using geometry (area of sector vs triangle), we can show this rigorously.
Derived Trigonometric Limits
$$\boxed{\lim_{x \to 0} \frac{\tan x}{x} = 1}$$ $$\boxed{\lim_{x \to 0} \frac{\sin^{-1} x}{x} = 1}$$ $$\boxed{\lim_{x \to 0} \frac{\tan^{-1} x}{x} = 1}$$ $$\boxed{\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}}$$ $$\boxed{\lim_{x \to 0} \frac{1 - \cos x}{x} = 0}$$General Forms
$$\boxed{\lim_{x \to 0} \frac{\sin(ax)}{x} = a}$$ $$\boxed{\lim_{x \to 0} \frac{\sin(ax)}{\sin(bx)} = \frac{a}{b}}$$ $$\boxed{\lim_{x \to 0} \frac{\tan(ax)}{\tan(bx)} = \frac{a}{b}}$$For $\lim_{x \to 0} \frac{\tan x}{x}$:
$$\frac{\tan x}{x} = \frac{\sin x}{x \cos x} = \frac{\sin x}{x} \cdot \frac{1}{\cos x}$$ $$\lim_{x \to 0} \frac{\tan x}{x} = \lim_{x \to 0} \frac{\sin x}{x} \cdot \lim_{x \to 0} \frac{1}{\cos x} = 1 \cdot 1 = 1$$Category 2: Exponential Limits
The Fundamental Exponential Limits
$$\boxed{\lim_{x \to 0} \frac{e^x - 1}{x} = 1}$$ $$\boxed{\lim_{x \to 0} \frac{a^x - 1}{x} = \ln a}$$ $$\boxed{\lim_{x \to 0} \frac{a^x - b^x}{x} = \ln a - \ln b = \ln\left(\frac{a}{b}\right)}$$The Famous $e$ Limit
$$\boxed{\lim_{x \to 0} (1 + x)^{1/x} = e}$$ $$\boxed{\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x = e}$$ $$\boxed{\lim_{x \to \infty} \left(1 + \frac{a}{x}\right)^x = e^a}$$Generalization:
$$\boxed{\lim_{x \to a} \left(1 + \frac{1}{f(x)}\right)^{f(x)} = e \text{ when } f(x) \to \infty}$$The limit $\lim_{n \to \infty} \left(1 + \frac{r}{n}\right)^n = e^r$ describes compound interest!
If you invest money at rate $r$ compounded continuously, your money grows by factor $e^r$. This is why banks love exponentials — and why you should love this limit!
Category 3: Logarithmic Limits
Fundamental Logarithmic Limits
$$\boxed{\lim_{x \to 0} \frac{\ln(1 + x)}{x} = 1}$$ $$\boxed{\lim_{x \to 0} \frac{\log_a(1 + x)}{x} = \frac{1}{\ln a}}$$Generalization:
$$\boxed{\lim_{x \to 0} \frac{\ln(1 + ax)}{x} = a}$$Notice the symmetry:
- $\lim_{x \to 0} \frac{e^x - 1}{x} = 1$ (exponential)
- $\lim_{x \to 0} \frac{\ln(1 + x)}{x} = 1$ (logarithm)
This is because $\ln$ and $e^x$ are inverse functions!
Category 4: Algebraic Limits
The Power Formula
$$\boxed{\lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1}}$$This is actually the derivative of $x^n$ from first principles!
Special cases:
$$\lim_{x \to 0} \frac{x^n}{x} = \begin{cases} 0 & \text{if } n > 1 \\ 1 & \text{if } n = 1 \\ \infty & \text{if } n < 1 \end{cases}$$Generalized Power Limit
$$\boxed{\lim_{x \to 0} \frac{(1 + x)^n - 1}{x} = n}$$Category 5: The $1^\infty$ Form
Master Formula for $1^\infty$
When $\lim f(x) = 1$ and $\lim g(x) = \infty$:
$$\boxed{\lim [f(x)]^{g(x)} = e^{\lim g(x)[f(x) - 1]}}$$Application: This solves all $1^\infty$ indeterminate forms!
Example:
$$\lim_{x \to \infty} \left(\frac{x + 3}{x + 1}\right)^{2x}$$Here $f(x) = \frac{x+3}{x+1} \to 1$ and $g(x) = 2x \to \infty$.
$$f(x) - 1 = \frac{x+3}{x+1} - 1 = \frac{2}{x+1}$$ $$g(x)[f(x) - 1] = 2x \cdot \frac{2}{x+1} = \frac{4x}{x+1} \to 4$$ $$\lim_{x \to \infty} \left(\frac{x + 3}{x + 1}\right)^{2x} = e^4$$L’Hôpital’s Rule (JEE Advanced)
The Rule
For indeterminate forms $\frac{0}{0}$ or $\frac{\infty}{\infty}$:
$$\boxed{\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}}$$Conditions:
- The limit gives $\frac{0}{0}$ or $\frac{\infty}{\infty}$
- Both $f$ and $g$ are differentiable near $a$
- The limit of $\frac{f'(x)}{g'(x)}$ exists
When to Use L’Hôpital
Use L’Hôpital when:
- You recognize $\frac{0}{0}$ or $\frac{\infty}{\infty}$ form
- Factorization/rationalization seems messy
- You’re allowed to use derivatives (JEE Advanced)
Use Standard Limits when:
- You recognize a standard pattern
- JEE Main problem (L’Hôpital not in syllabus)
- Faster than differentiating
Examples Using L’Hôpital
Example 1:
$$\lim_{x \to 0} \frac{\sin x - x}{x^3}$$Direct: $\frac{0}{0}$ form.
Apply L’Hôpital:
$$= \lim_{x \to 0} \frac{\cos x - 1}{3x^2} \quad (\text{still } \frac{0}{0})$$Apply again:
$$= \lim_{x \to 0} \frac{-\sin x}{6x} = \frac{-1}{6} \lim_{x \to 0} \frac{\sin x}{x} = \frac{-1}{6}$$Example 2:
$$\lim_{x \to \infty} \frac{\ln x}{x}$$Form: $\frac{\infty}{\infty}$
$$= \lim_{x \to \infty} \frac{1/x}{1} = \lim_{x \to \infty} \frac{1}{x} = 0$$Mistake 1: Using L’Hôpital when form is NOT $\frac{0}{0}$ or $\frac{\infty}{\infty}$
Mistake 2: Using quotient rule instead of differentiating numerator and denominator separately
Wrong: $\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}$ ✗
Correct for L’Hôpital: $\frac{f'(x)}{g'(x)}$ ✓
Memory Tricks & Patterns
The “SET” Mnemonic
Standard limits fall into three categories — remember “SET”:
Sin limits: $\frac{\sin x}{x} = 1$, $\frac{1 - \cos x}{x^2} = \frac{1}{2}$
Exponential limits: $\frac{e^x - 1}{x} = 1$, $(1 + x)^{1/x} = e$
Tangent & others: $\frac{\tan x}{x} = 1$, $\frac{\ln(1+x)}{x} = 1$
The Symmetry Pattern
Notice the beautiful symmetry:
| Exponential | Logarithmic |
|---|---|
| $\frac{e^x - 1}{x} = 1$ | $\frac{\ln(1 + x)}{x} = 1$ |
| $\frac{a^x - 1}{x} = \ln a$ | $\frac{\log_a(1 + x)}{x} = \frac{1}{\ln a}$ |
They’re inverses of each other!
The “Magic $e$” Family
All these equal $e$:
$$\lim_{x \to 0} (1 + x)^{1/x} = e$$ $$\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x = e$$ $$\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e$$Memory trick: “Everyone Approaches 1 but gets raised to Infinity”
Quick Reference Table
All Standard Limits at a Glance
| Limit | Value |
|---|---|
| $\lim_{x \to 0} \frac{\sin x}{x}$ | $1$ |
| $\lim_{x \to 0} \frac{\tan x}{x}$ | $1$ |
| $\lim_{x \to 0} \frac{1 - \cos x}{x^2}$ | $\frac{1}{2}$ |
| $\lim_{x \to 0} \frac{e^x - 1}{x}$ | $1$ |
| $\lim_{x \to 0} \frac{a^x - 1}{x}$ | $\ln a$ |
| $\lim_{x \to 0} \frac{\ln(1 + x)}{x}$ | $1$ |
| $\lim_{x \to 0} (1 + x)^{1/x}$ | $e$ |
| $\lim_{x \to \infty} (1 + 1/x)^x$ | $e$ |
| $\lim_{x \to a} \frac{x^n - a^n}{x - a}$ | $na^{n-1}$ |
| $\lim_{x \to 0} \frac{(1+x)^n - 1}{x}$ | $n$ |
Generalized Forms
| Pattern | Formula |
|---|---|
| $\lim_{x \to 0} \frac{\sin(ax)}{bx}$ | $\frac{a}{b}$ |
| $\lim_{x \to 0} \frac{e^{ax} - 1}{bx}$ | $\frac{a}{b}$ |
| $\lim_{x \to 0} \frac{\ln(1 + ax)}{bx}$ | $\frac{a}{b}$ |
| $\lim [f(x)]^{g(x)}$ when $f \to 1, g \to \infty$ | $e^{\lim g(x)[f(x)-1]}$ |
Practice Problems
Level 1: Foundation (NCERT)
Question: Evaluate $\lim_{x \to 0} \frac{\sin 5x}{x}$
Solution:
$$\lim_{x \to 0} \frac{\sin 5x}{x} = \lim_{x \to 0} \frac{\sin 5x}{5x} \cdot 5 = 1 \cdot 5 = 5$$Using $\lim_{x \to 0} \frac{\sin(ax)}{x} = a$ directly: Answer = $5$
Question: Find $\lim_{x \to 0} \frac{e^{3x} - 1}{x}$
Solution:
$$\lim_{x \to 0} \frac{e^{3x} - 1}{x} = \lim_{x \to 0} \frac{e^{3x} - 1}{3x} \cdot 3 = 1 \cdot 3 = 3$$Using $\lim_{x \to 0} \frac{e^{ax} - 1}{x} = a$ directly: Answer = $3$
Level 2: JEE Main
Question: Evaluate $\lim_{x \to 0} \frac{\sin 3x}{\tan 2x}$
Solution:
$$\frac{\sin 3x}{\tan 2x} = \frac{\sin 3x}{\sin 2x} \cdot \cos 2x$$ $$= \frac{\sin 3x}{3x} \cdot \frac{2x}{\sin 2x} \cdot \frac{3x}{2x} \cdot \cos 2x$$ $$\lim_{x \to 0} \left[1 \cdot 1 \cdot \frac{3}{2} \cdot 1\right] = \frac{3}{2}$$Question: Find $\lim_{x \to 0} \frac{2^x - 3^x}{x}$
Solution:
$$\lim_{x \to 0} \frac{2^x - 3^x}{x} = \lim_{x \to 0} \frac{2^x - 1}{x} - \lim_{x \to 0} \frac{3^x - 1}{x}$$ $$= \ln 2 - \ln 3 = \ln\left(\frac{2}{3}\right)$$Question: Evaluate $\lim_{x \to 0} \frac{1 - \cos 2x}{x^2}$
Solution:
$$\lim_{x \to 0} \frac{1 - \cos 2x}{x^2} = \lim_{x \to 0} \frac{1 - \cos 2x}{(2x)^2} \cdot 4 = \frac{1}{2} \cdot 4 = 2$$Using $\lim_{x \to 0} \frac{1 - \cos(ax)}{x^2} = \frac{a^2}{2}$ directly: $\frac{4}{2} = 2$
Level 3: JEE Advanced
Question: Find $\lim_{x \to \infty} \left(\frac{x + 5}{x + 3}\right)^{x + 2}$
Solution: This is $1^\infty$ form.
Let $f(x) = \frac{x+5}{x+3}$ and $g(x) = x + 2$.
$$f(x) - 1 = \frac{x+5}{x+3} - 1 = \frac{2}{x+3}$$ $$g(x)[f(x) - 1] = (x+2) \cdot \frac{2}{x+3} = \frac{2x + 4}{x + 3}$$ $$\lim_{x \to \infty} \frac{2x + 4}{x + 3} = 2$$ $$\lim_{x \to \infty} \left(\frac{x + 5}{x + 3}\right)^{x + 2} = e^2$$Question: Evaluate $\lim_{x \to 0} \frac{x - \sin x}{x^3}$
Solution: Form: $\frac{0}{0}$
Apply L’Hôpital:
$$= \lim_{x \to 0} \frac{1 - \cos x}{3x^2} \quad (\text{still } \frac{0}{0})$$Apply again:
$$= \lim_{x \to 0} \frac{\sin x}{6x} = \frac{1}{6} \cdot 1 = \frac{1}{6}$$Question: Find $\lim_{x \to 0} \left(\frac{\sin x}{x}\right)^{1/x^2}$
Solution: This is $1^\infty$ form.
Let $f(x) = \frac{\sin x}{x}$ and $g(x) = \frac{1}{x^2}$.
Using Taylor series: $\sin x = x - \frac{x^3}{6} + \ldots$
$$f(x) = \frac{x - x^3/6 + \ldots}{x} = 1 - \frac{x^2}{6} + \ldots$$ $$f(x) - 1 = -\frac{x^2}{6}$$ $$g(x)[f(x) - 1] = \frac{1}{x^2} \cdot \left(-\frac{x^2}{6}\right) = -\frac{1}{6}$$ $$\lim_{x \to 0} \left(\frac{\sin x}{x}\right)^{1/x^2} = e^{-1/6}$$Common Mistakes to Avoid
Wrong:
$$\lim_{x \to 0} \frac{\sin 3x}{2x} = 1$$✗
Correct:
$$\lim_{x \to 0} \frac{\sin 3x}{2x} = \lim_{x \to 0} \frac{\sin 3x}{3x} \cdot \frac{3}{2} = 1 \cdot \frac{3}{2} = \frac{3}{2}$$✓
You must have $\frac{\sin(ax)}{ax}$ to get 1!
Wrong:
$$\lim_{x \to 0} \frac{2^x - 1}{x} = 1$$✗
Correct:
$$\lim_{x \to 0} \frac{2^x - 1}{x} = \ln 2$$✓
Only $e^x$ gives 1, others give $\ln a$!
Wrong:
$$\lim_{x \to 0} \frac{x + 1}{x + 2} \stackrel{?}{=} \lim_{x \to 0} \frac{1}{1} = 1$$✗
This gives $\frac{1}{2}$, not $\frac{0}{0}$! Don’t use L’Hôpital.
Correct: Direct substitution: $\frac{0 + 1}{0 + 2} = \frac{1}{2}$ ✓
Quick Revision Box
| Form | Standard Limit to Use |
|---|---|
| $\frac{\sin(ax)}{x}$ | Equals $a$ |
| $\frac{e^{ax} - 1}{x}$ | Equals $a$ |
| $\frac{\ln(1 + ax)}{x}$ | Equals $a$ |
| $\frac{1 - \cos(ax)}{x^2}$ | Equals $\frac{a^2}{2}$ |
| $(1 + x)^{1/x}$ as $x \to 0$ | Equals $e$ |
| $(1 + 1/x)^x$ as $x \to \infty$ | Equals $e$ |
| $f(x)^{g(x)}$ when $f \to 1, g \to \infty$ | $e^{\lim g(x)[f(x)-1]}$ |
JEE Strategy Tips
Speed Hack: If you see $\sin$, $\tan$, $e^x$, or $\ln$ in a limit, immediately think “standard limits” — it’s probably a 30-second problem!
JEE Main: Focus on direct application of standard limits. L’Hôpital’s Rule is NOT in the syllabus.
JEE Advanced: Master both standard limits AND L’Hôpital. Advanced problems combine multiple techniques.
Common Pattern: JEE loves $1^\infty$ forms. Practice the formula $e^{\lim g(x)[f(x)-1]}$ until it’s muscle memory.
Time Saver: Make flash cards for all standard limits. Recognition speed is everything!
Teacher’s Summary
Standard limits are your first weapon against difficult limit problems. Always check if a standard pattern applies before trying other methods.
The three pillars: Sin/Tan limits (value 1), Exponential limits ($\ln a$), and the magic $e$ family. Master these and you’ll solve 70% of problems instantly.
L’Hôpital is powerful but has conditions — only for $\frac{0}{0}$ or $\frac{\infty}{\infty}$, and differentiate separately, NOT using quotient rule.
The $1^\infty$ formula is your JEE Advanced secret weapon: $\lim [f(x)]^{g(x)} = e^{\lim g(x)[f(x)-1]}$
Symmetry is beautiful: Notice how $e^x$ and $\ln x$ limits mirror each other — they’re inverse functions!
“Standard limits turn impossible problems into one-line solutions!”
Related Topics
Within Limits, Continuity & Differentiability
- Introduction to Limits — Foundation concepts
- Continuity — Using limits to check continuity
- Differentiability — Derivatives use standard limits
Mathematical Foundations
- Exponential Functions — Understanding $e^x$ behavior
- Trigonometry — Sin, cos, tan functions
- Logarithms — Properties of $\ln$ and $\log$
Applications
- Derivatives & Differentiation — Standard limits derive basic derivatives
- Integral Calculus — Limits define integrals
- Series & Sequences — $e$ defined as series limit
Physics Connections
- Kinematics — Instantaneous velocity from limits
- Simple Harmonic Motion — Sin/cos limits in oscillations
- Radioactive Decay — Exponential limit applications