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$f(x)=\int_{0}^{x}\cos(1/t)dt$

Then$f'(0)=?$

I tried $f'(x)=\cos\left(1/x\right)$, but then I get into trouble.

Hope for your help

skyking
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z3wood
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2 Answers2

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You need to go back to the definition: $$f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}{x}=\lim_{x\to0}\frac1x\int_0^x\cos(1/t)\,dt.$$ Now the integrand oscillates like crazy near the origin, and it's not unreasonable to guess that the limit will be zero. To prove it, I suggest the substitution $t=1/s$ in the integral, then use $\lvert\cos s\rvert\le1$. Can you take it from there?

(Edit: Fixed typo in the very first formula.)

Harald Hanche-Olsen
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You could try integration by parts: $$ \int_0^x\cos(1/t)dt=-t^2\sin(1/t)|_0^x+2\int_0^xt\sin(1/t)dt. $$ Both terms are $O(x^2)$.

user254433
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  • shouldn't it be $$ \int_0^x\cos(1/t)dt=t\cos(1/t)|_0^x+\int_0^x \frac{1}{t}\sin(1/t)dt. $$ –  Jul 21 '17 at 10:30
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    @Bacon I integrated the cosine not 1: $$\cos(1/t) = -t^2\frac{\cos(1/t)}{-t^2}=-t^2\frac{d}{dt}\sin(1/t)$$. – user254433 Jul 21 '17 at 10:42
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    After I posted that I thought perhaps you'd chosen a different $dv$ in the partial integration. My mistake - Thanks! –  Jul 21 '17 at 11:42