---
title: "Limit of (sin x / x)^(1/x)"
subject: "Mathematics I"
chapter: 1
key: "semester_1/mathematics_i:limit-sinx-over-x-power-1-over-x"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/limit-sinx-over-x-power-1-over-x
site: BCT Engineering
---

# Limit of (sin x / x)^(1/x)

Mathematics I · Chapter 1: Derivatives and their Applications

- Asked in: 2073 Shrawan; 2067 Asar (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/limit-sinx-over-x-power-1-over-x
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-1

## Question

Evaluate lim_{x→0⁺} (sin x / x)^(1/x).

## Short answer

Let L = lim_{x→0⁺} (sin x/x)^{1/x}. Take log: ln L = lim ln(sin x/x)/x. Using series or L'Hôpital, the limit is 0, so L = e⁰ = 1.

## Answer

Let L = lim_{x→0⁺} (sin x / x)^(1/x). As x → 0, sin x / x → 1 and 1/x → ∞, so this is the form 1^∞.

Take logarithms:

ln L = lim_{x→0⁺} ln(sin x / x) / x = lim [ln sin x − ln x] / x  (form 0/0)

Apply L'Hospital's rule:

```text
ln L = lim [ cot x − 1/x ]
     = lim (x cos x − sin x) / (x sin x)      (0/0)
     = lim (−x sin x) / (sin x + x cos x)     (differentiate top and bottom)
     = 0 / 0 → differentiate again:
       lim (−sin x − x cos x) / (2 cos x − x sin x) = 0 / 2 = 0
```
Check by series: sin x / x = 1 − x²/6 + ..., so ln(sin x/x) ≈ −x²/6, and (−x²/6)/x = −x/6 → 0.

So ln L = 0.

**Answer: L = e⁰ = 1.**

## Check yourself

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

1. (sin x / x)^(1/x) as x tends to 0+ is of the form:
   A. 1^infinity
   B. 0^0
   C. infinity^0
   D. 0/0
   Answer: A. Base tends to 1, exponent to infinity. (Key point: Form)

2. Taking logs and applying L'Hospital gives lim [cot x - 1/x]. This is rewritten as:
   A. (x cos x - sin x)/(x sin x)
   B. (x - sin x)/x
   C. sin x/x
   D. (cos x - 1)/x
   Answer: A. cot x - 1/x = (x cos x - sin x)/(x sin x), form 0/0. (Key point: After logs)

3. ln(sin x/x) is approximately -x^2/6, so divided by x it tends to:
   A. -infinity
   B. 1
   C. -1/6
   D. 0
   Answer: D. -x/6 tends to 0. (Key point: Series check)

4. The limit L is:
   A. e^(-1/6)
   B. 1
   C. e
   D. 0
   Answer: B. ln L = 0 so L = 1. (Key point: Result)

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