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Exercise 707Calculus › Single-variable calculus › LimitsL2Added 1 Nov 2026

Cosine to the power 1/x²

#exponential form#Taylor expansion#indeterminate forms

Statement

Compute
lim⁡x→0(cos⁡x)1/x2.\lim_{x\to0}(\cos x)^{1/x^2}.

Hint

Write it as exp⁡ ⁣(ln⁡cos⁡xx2)\exp\!\left(\frac{\ln\cos x}{x^2}\right) and expand cos⁡x\cos x, then ln⁡(1+h)\ln(1+h).

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Solution

Idea. Another "1∞1^\infty" form: the base tends to 11 and the exponent to +∞+\infty. We take the logarithm to turn the power into a product.

Computation. For xx near 00, cos⁡x>0\cos x>0 and
(cos⁡x)1/x2=exp⁡ ⁣(ln⁡cos⁡xx2).(\cos x)^{1/x^2}=\exp\!\left(\frac{\ln\cos x}{x^2}\right).
Write cos⁡x=1+h\cos x=1+h with h=−x22+o(x2)h=-\frac{x^2}2+o(x^2). Then h→0h\to0 and
ln⁡cos⁡x=ln⁡(1+h)=h+o(h)=−x22+o(x2).\ln\cos x=\ln(1+h)=h+o(h)=-\frac{x^2}2+o(x^2).
Hence ln⁡cos⁡xx2→−12\dfrac{\ln\cos x}{x^2}\to-\dfrac12, and by continuity of exp⁡\exp the limit is e−1/2=1ee^{-1/2}=\dfrac1{\sqrt e}.

Remark. The pattern is general: if u(x)→1u(x)\to1, then u(x)1/x2u(x)^{1/x^2} is governed by the limit of u(x)−1x2\dfrac{u(x)-1}{x^2}.