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let $$f(x)=\int_{0}^{x}\cos{\dfrac{1}{t}}dt$$, Find the $f'(0)$

since $$f'(0)=\lim_{x\to 0}\dfrac{f(x)-f(0)}{x}=\lim_{x\to 0}\dfrac{\int_{0}^{x}\cos{\dfrac{1}{t}}dt}{x}$$ since By Partial integration yields $$\int_{0}^{x}\cos{\dfrac{1}{t}}dt=\int_{\frac{1}{x}}^{+\infty}\dfrac{\cos{u}}{u^2}du=-x^2\sin{\dfrac{1}{x}}+\int_{\frac{1}{x}}^{\infty}\dfrac{2\sin{u}}{u^3}du$$ so $$\left|\dfrac{\int_{0}^{x}\cos{\frac{1}{t}}dt}{x}\right|\le|x|\left|\sin{\dfrac{1}{x}}\right|+\dfrac{1}{|x|}\int_{\frac{1}{x}}^{\infty}\dfrac{2}{u^3}=|x|\left|\sin{\dfrac{1}{x}}\right|+|x|\to 0(x\to 0)$$

My Question: This problem have other methods?

Thank you

math110
  • 93,304

1 Answers1

2

You can also just integrate by parts in the integral directly: $$f(x) = \int_0^x (-t^2){d \over dt} \bigg(\sin {1 \over t}\bigg)\,dt$$ $$= -x^2\sin{1 \over x} + \int_0^{x} 2t\cos\bigg({1 \over t}\bigg)\,dt$$ (The limit as $x$ goes to zero of the boundary term is just zero.) The absolute value of $2t\cos\big({1 \over t}\big)$ is at most $2|t|$, so the second term is at most $x^2$. Hence $\big|{f(x) \over x}\big| \leq 2|x|$ and therefore the limit of the difference quotients is zero.

Of course this isn't that different from what you were doing, but you don't have to do any variable changes this way.

Zarrax
  • 44,950