Sandwich Theorem (Squeeze Theorem): Mastering the Technique
Learn to bound functions, prove limits exist, and solve complex limit problems with this powerful JEE tool.
What is the Sandwich Theorem?
The Sandwich Theorem (also called Squeeze Theorem) is a fundamental tool in calculus used to evaluate limits of functions that are difficult to compute directly. It states that if a function $f(x)$ is "squeezed" between two other functions $g(x)$ and $h(x)$ that approach the same limit $L$ as $x$ approaches $c$, then $f(x)$ must also approach $L$.
Formal Statement
If $g(x) \leq f(x) \leq h(x)$ for all $x$ in some interval around $c$ (except possibly at $c$ itself), and
$\lim_{x \to c} g(x) = \lim_{x \to c} h(x) = L$
Then $\lim_{x \to c} f(x) = L$
Bounding Trigonometric Functions
The most common application: using $-1 \leq \sin x \leq 1$ and $-1 \leq \cos x \leq 1$
Example: Find $\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right)$
Step 1: We know $-1 \leq \sin\left(\frac{1}{x}\right) \leq 1$ for all $x \neq 0$
Step 2: Multiply by $x^2$ (positive for $x \neq 0$): $-x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2$
Step 3: Find limits of bounding functions:
• $\lim_{x \to 0} (-x^2) = 0$
• $\lim_{x \to 0} x^2 = 0$
Step 4: By Sandwich Theorem: $\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0$
Proving Limits Exist
When direct substitution fails, Sandwich Theorem can prove a limit exists.
Example: Prove $\lim_{x \to 0} \frac{\sin x}{x} = 1$
Step 1: For small $x > 0$, geometric argument shows:
$\sin x \leq x \leq \tan x$
Step 2: Divide by $\sin x$: $1 \leq \frac{x}{\sin x} \leq \frac{1}{\cos x}$
Step 3: Take reciprocals: $\cos x \leq \frac{\sin x}{x} \leq 1$
Step 4: As $x \to 0$, $\cos x \to 1$ and $1 \to 1$
Step 5: By Sandwich Theorem: $\lim_{x \to 0} \frac{\sin x}{x} = 1$
Complex Oscillatory Functions
Handling functions that oscillate infinitely near a point.
Example: Find $\lim_{x \to 0} x^3 \cos\left(\frac{1}{x^2}\right)$
Step 1: Bound the cosine: $-1 \leq \cos\left(\frac{1}{x^2}\right) \leq 1$
Step 2: Multiply by $x^3$: $-x^3 \leq x^3 \cos\left(\frac{1}{x^2}\right) \leq x^3$
Step 3: Evaluate limits:
• $\lim_{x \to 0} (-x^3) = 0$
• $\lim_{x \to 0} x^3 = 0$
Step 4: By Sandwich Theorem: $\lim_{x \to 0} x^3 \cos\left(\frac{1}{x^2}\right) = 0$
🚀 Problem-Solving Strategy
When to Use Sandwich Theorem:
- Function oscillates (sine, cosine of $\frac{1}{x}$ etc.)
- Direct substitution gives indeterminate form
- Function is bounded between simpler functions
- Proving limit exists without finding exact value
Key Steps:
- Identify the oscillating/bounded part
- Find upper and lower bounds
- Multiply/divide to isolate the function
- Check if bounding functions have same limit
- Apply the theorem
Advanced Applications Available
Includes 3 more complex applications with step-by-step proofs and JEE Advanced level problems
📝 Quick Self-Test
Try these JEE-level problems using Sandwich Theorem:
1. Find $\lim_{x \to 0} x^4 \sin\left(\frac{1}{\sqrt[3]{x}}\right)$
2. Evaluate $\lim_{x \to \infty} \frac{\sin x}{x}$
3. Find $\lim_{x \to 0} \frac{1 - \cos x}{x^2}$ using bounding
⚠️ Common JEE Mistakes to Avoid
Forgetting to check bounds
Always verify $g(x) \leq f(x) \leq h(x)$ in the interval
Using wrong inequality direction
When multiplying by negative numbers, reverse inequalities
Not verifying bounding limits match
Both bounding functions must approach the same limit
Ready to Master Sandwich Theorem?
Get complete access to all applications with step-by-step video solutions and JEE practice problems