Lim f'(x)=0 implies lim f(x)/x = 0

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Homework Statement


Given that the function f is differentiable on the interval (a,\infty) and that \lim_{x\rightarrow\infty}f'(x)= 0. Show that \lim_{x\rightarrow\infty}\frac{f(x)}{x}= 0.

Homework Equations


The Attempt at a Solution


I have a pretty good intuition of why this is true: As f(x) approaches infinity, its derivative approaches 0. This means that, at some point, the function slope becomes less than that of x.
This means that when x approaches infinity, x is dominant over f(x) and thus f(x)/x => 0.

I can't come up with any strict mathematical proof though, so I would be very greatful if someone could help me with this.
 
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Kreutzfelt said:

Homework Statement


Given that the function f is differentiable on the interval (a,\infty) and that \lim_{x\rightarrow\infty}f'(x)= 0. Show that \lim_{x\rightarrow\infty}\frac{f(x)}{x}= 0.

Homework Equations





The Attempt at a Solution


I have a pretty good intuition of why this is true: As f(x) approaches infinity, its derivative approaches 0. This means that, at some point, the function slope becomes less than that of x.
This means that when x approaches infinity, x is dominant over f(x) and thus f(x)/x => 0.

I can't come up with any strict mathematical proof though, so I would be very greatful if someone could help me with this.

L' Hopital's Rule?
 
L'Hopital's rule is unfortunately not allowed in this particular problem.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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