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

# Limit of ((1+x)^(1/x) − e)/x

Mathematics I · Chapter 1: Derivatives and their Applications

- Asked in: 2072 Chaitra
- 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-plus-x-power-1-over-x-minus-e
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-1

## Question

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

## Short answer

Use series expansion: (1+x)^(1/x) = e^(ln(1+x)/x) = e·e^(−x/2+…) = e(1 − x/2 + …). Then the limit becomes −e/2.

## Answer

Let L = lim_{x→0} [(1 + x)^(1/x) − e]/x. This is 0/0 since (1+x)^(1/x) → e.

Write (1 + x)^(1/x) = e^(ln(1+x)/x). Expand:

ln(1 + x)/x = 1 − x/2 + x²/3 − ...

(1 + x)^(1/x) = e^(1 − x/2 + x²/3 − ...) = e · e^(−x/2 + x²/3 − ...) = e[1 − x/2 + O(x²)]

So (1 + x)^(1/x) − e = −e x/2 + O(x²), and dividing by x:

**Answer: L = −e/2.**

(By L'Hospital: with y = (1+x)^(1/x), y′ = y[1/(x(1+x)) − ln(1+x)/x²]. The bracket tends to −1/2 as x → 0, giving y′ → −e/2.)

## Check yourself

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

1. ((1+x)^(1/x) - e)/x is of the form:
   A. 0/0
   B. 1^infinity
   C. infinity/infinity
   D. 0 * infinity
   Answer: A. (1+x)^(1/x) tends to e. (Key point: Form)

2. (1+x)^(1/x) is written as:
   A. e^(ln(1+x)/x)
   B. e^(x ln(1+x))
   C. e^(1/x)
   D. x^(1/x)
   Answer: A. Exponential-log form. (Key point: Rewrite)

3. ln(1+x)/x =
   A. x - x^2/2
   B. 1 - x/2 + x^2/3 - ...
   C. 1 - x + x^2
   D. 1 + x/2 + x^2/3
   Answer: B. Series of ln(1+x) divided by x. (Key point: Expansion)

4. The limit is:
   A. -e/2
   B. e/2
   C. e
   D. 0
   Answer: A. e(-x/2)/x = -e/2. (Key point: Result)

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