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Exercise 33Calculus › Single-variable calculus › Differentiation › Taylor expansionsL2Added 3 Nov 2026

lim ((1 + x)^(1/x) / e)^(1/x)

#limits#indeterminate form 1^∞#logarithm#expansions

Statement

Compute lim⁡x→0((1+x)1/xe)1/x\displaystyle\lim_{x\to0}\left(\frac{(1+x)^{1/x}}e\right)^{1/x}.

Hint

Take the logarithm. You need ln⁡(1+x)\ln(1+x) up to order 22.

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Solution

Logarithm. Let L(x)=1xln⁡(1+x)1/xe=1x(ln⁡(1+x)x−1)L(x)=\frac1x\ln\frac{(1+x)^{1/x}}e=\frac1x\left(\frac{\ln(1+x)}x-1\right).

Expansion. ln⁡(1+x)=x−x22+x33+o(x3)\ln(1+x)=x-\frac{x^2}2+\frac{x^3}3+o(x^3), so
ln⁡(1+x)x−1=−x2+x23+o(x2).\frac{\ln(1+x)}x-1=-\frac x2+\frac{x^2}3+o(x^2).

Limit. Therefore L(x)=−12+x3+o(x)→−12L(x)=-\frac12+\frac x3+o(x)\to-\frac12, and by continuity of exp⁡\exp the limit is
e−1/2=1e.e^{-1/2}=\frac1{\sqrt e}.

Remark. This also shows that (1+x)1/x=e(1−x2+o(x))(1+x)^{1/x}=e\left(1-\frac x2+o(x)\right): it approaches ee linearly, with slope −e2-\frac e2. (Compare with the expansion of (1+1n)n(1+\frac1n)^n.)