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Exercise 175Calculus › Single-variable calculus › Differentiation › Optimisation and curve sketchingL1Added 4 Nov 2026

Complete study of (ln x)/x

#curve sketching#asymptotes#inflection point#logarithm

Statement

Study f(x)=ln⁡xxf(x)=\dfrac{\ln x}x on (0,+∞)(0,+\infty): limits, asymptotes, variations, convexity, graph. Deduce that ln⁡x≤xe\ln x\le\dfrac xe for all x>0x>0.

Hint

f′(x)=1−ln⁡xx2f'(x)=\dfrac{1-\ln x}{x^2}. For the second derivative, differentiate the quotient again.

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Solution

Limits. As x→0+x\to0^+: ln⁡x→−∞\ln x\to-\infty, 1x→+∞\frac1x\to+\infty, so f→−∞f\to-\infty: the line x=0x=0 is a vertical asymptote. As x→+∞x\to+\infty: f→0f\to0 (growth comparison), so y=0y=0 is a horizontal asymptote. The only zero is x=1x=1.

Variations. f′>0f'>0 for x<ex<e and f′<0f'<0 for x>ex>e: increasing on (0,e](0,e], decreasing on [e,∞)[e,\infty). Maximum f(e)=1ef(e)=\frac1e.

Convexity. f′′(x)=2ln⁡x−3x3f''(x)=\dfrac{2\ln x-3}{x^3}: concave on (0,e3/2](0,e^{3/2}], convex on [e3/2,∞)[e^{3/2},\infty), inflection point at x=e3/2≈4.48x=e^{3/2}\approx4.48, where f=32e3/2≈0.335f=\frac3{2e^{3/2}}\approx0.335.

The inequality. f(x)≤f(e)=1ef(x)\le f(e)=\frac1e, i.e. ln⁡x≤xe\ln x\le\frac xe for x>0x>0 (equality at x=ex=e). It is the tangent line of ln⁡\ln at ee.

Remark. f(a)=f(b)f(a)=f(b) with a<ba<b forces a<e<ba<e<b. For positive integers, the only pair a<ba<b with ab=baa^b=b^a is (2,4)(2,4), as f(2)=f(4)=ln⁡22f(2)=f(4)=\frac{\ln2}2 (indeed 24=42=162^4=4^2=16). (For negative integers, ab=baa^b=b^a is not equivalent to f(a)=f(b)f(a)=f(b), since ln⁡\ln is not defined there.)