6. limx → 0 sin(x) / x is __________.
Related Keywords: sinc function limit, standard trigonometric limit, indeterminate form 0/0, calculus proof
Correct Answer
1
Why Direct Substitution Fails
Substituting x = 0 gives:
sin(0)/0 = 0/0
This is an indeterminate form, so we need another method to evaluate the limit.
Explanations of Common Multiple-Choice Options
| Option | Meaning | Correct/Incorrect | Explanation |
|---|---|---|---|
| 0 | Common misconception | ❌ Incorrect | Both numerator and denominator tend to 0, but their ratio approaches 1. |
| 1/2 | Distractor based on geometry | ❌ Incorrect | Appears in other limits but not this one. |
| Undefined | Division by zero | ❌ Incorrect | The function is undefined at x = 0, but the limit exists. |
| 1 | Standard calculus fact | ✅ Correct | Confirmed by multiple proof methods. |
Methods to Prove the Limit
1. L’Hôpital’s Rule
The limit is 0/0, so differentiate top and bottom:
sin(x) → cos(x) and x → 1
lim (x → 0) cos(x)/1 = 1
2. Squeeze Theorem
For small x in radians:
sin(x) < x < tan(x)
Dividing through by sin(x) gives:
cos(x) < sin(x)/x < 1
As x → 0, cos(x) → 1, so the squeezed middle term → 1.
3. Taylor Series
sin(x) = x - x³/6 + O(x⁵)
Dividing by x:
sin(x)/x = 1 - x²/6 + O(x⁴) → 1
Common Misconceptions
- Thinking sin(x) ≈ 0 means the quotient tends to 0
- Confusing the function value with the limit
- Not using radians
Key Takeaway
The limit:
lim (x → 0) sin(x)/x = 1
This result underpins:
- Trigonometric derivatives
- Fourier analysis
- Sinc function behavior
- Taylor expansions
Final Answer: 1
