---
title: "Limit of 1/x² − 1/sin²x"
subject: "Mathematics I"
chapter: 1
key: "semester_1/mathematics_i:limit-1-over-x2-minus-1-over-sin2"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/limit-1-over-x2-minus-1-over-sin2
site: BCT Engineering
---

# Limit of 1/x² − 1/sin²x

Mathematics I · Chapter 1: Derivatives and their Applications

- Asked in: 2069 Asar; 2068 Chaitra (asked 2 times)
- Video: [Evaluate || lim_(𝒙→𝟎)⁡[𝟏/𝒙^𝟐 −𝟏/((𝒔𝒊𝒏)^𝟐 𝒙)] || Differential Calculus ||](https://www.youtube.com/watch?v=Rc4xXVKQWpo) — Math E Connect
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/limit-1-over-x2-minus-1-over-sin2
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-1

## Question

Evaluate lim_{x→0} (1/x² − 1/sin²x).

## Short answer

Combine fractions: 1/x²−1/sin²x = (sin²x−x²)/(x² sin²x). Using sin x = x−x³/6+..., numerator ~−x⁴/3, denominator ~x⁴, so the limit is −1/3.

## Answer

The limit is of the form ∞ − ∞. Combine the fractions:

1/x² − 1/sin²x = (sin²x − x²) / (x² sin²x)  (form 0/0)

Use the series sin x = x − x³/6 + x⁵/120 − ...

```text
sin²x = x² − x⁴/3 + 2x⁶/45 − ...
sin²x − x² = −x⁴/3 + 2x⁶/45 − ...
x² sin²x   = x⁴ − x⁶/3 + ...
```
Divide numerator and denominator by x⁴:

(−1/3 + 2x²/45 − ...) / (1 − x²/3 + ...) → −1/3 as x → 0.

(Same result by L'Hospital's rule applied four times, since both numerator and denominator vanish to order x⁴.)

**Answer: −1/3.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. 1/x^2 - 1/sin^2 x is of the form:
   A. 0 * infinity
   B. 0/0
   C. 1^infinity
   D. infinity - infinity
   Answer: D. Both terms blow up. (Key point: Form)

2. The first step is to:
   A. Take logs
   B. Combine into (sin^2 x - x^2)/(x^2 sin^2 x)
   C. Multiply by x
   D. Square the terms
   Answer: B. A common denominator turns it into 0/0. (Key point: Conversion)

3. sin^2 x - x^2 begins with:
   A. -x^4/3
   B. -x^2/3
   C. -x^6/3
   D. x^4/3
   Answer: A. sin^2 x = x^2 - x^4/3 + ... (Key point: Series of sin^2)

4. The limit is:
   A. -2/45
   B. 1/3
   C. 0
   D. -1/3
   Answer: D. (-x^4/3)/(x^4) gives -1/3. (Key point: Result)

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