I want to evaluate this limit:

$$\lim\limits_{x\to 0}\frac{1}{x}\int_{0}^{x}\sin^{2}\left(\frac{1}{u}\right)du$$

Here's my trial!

$$\lim\limits_{x\to 0}\frac{1}{x}\int_{0}^{x}\sin^{2}\left(\frac{1}{u}\right)du$$ $$=\lim\limits_{x\to 0}\frac{1}{2x}\int_{0}^{x}\left[1-\cos\left(\frac{2}{u}\right)\right]du$$ but I don't know how to proceed from here! Please, can someone help me out?


For the hard part, change variables with $t = 2/u, \, du = (-2/t^2) dt$ and integrate by parts to get

$$\frac{1}{2x}\int_0^x \cos(2/u) \, du = \frac{1}{x}\int_{2/x}^\infty \frac{\cos t}{t^2} \, dt \\ = -\frac{x\sin(2/x)}{4} + \frac{2}{x}\int_{2/x}^\infty\frac{\sin t}{t^3} \, dt \\ $$

Now you can take the limit as $x \to 0$ using

$$\left|\frac{1}{x}\int_{2/x}^\infty\frac{\sin t}{t^3} \, dt\right| \leqslant \frac{1}{x}\int_{2/x}^\infty\frac{1}{t^3} = \frac{x}{8}$$

showing that

$$\lim_{x \to 0} \frac{1}{2x}\int_0^x \cos(2/u) \, du = 0$$

  • $\begingroup$ $\frac{1}{2x}\int_0^x \cos(2/u) \, du = \frac{1}{2x}\int_{2/x}^\infty \frac{\cos t}{t^2} \, dt $. Please, kindly explain how you got this line! I mean transforming from $2/x$ to $\infty.$ $\endgroup$ – Omojola Micheal Aug 3 '18 at 20:27
  • $\begingroup$ @Mike: Let $t = 2/u$. Then $u = 2/t$ and $du = -2 dt /t^2$ the limits transform as $0 \to \infty$ and $x \to 2/x$. I may be off by a factor of two in the answer so I will edit. $\endgroup$ – RRL Aug 3 '18 at 20:32
  • $\begingroup$ Thanks, I will be waiting! $\endgroup$ – Omojola Micheal Aug 3 '18 at 20:34
  • $\begingroup$ The point is the entire expression is $\mathcal{O}(x)$ and so goes to $0$ as $x$ goes to $0$. Integration by parts reveals this. $\endgroup$ – RRL Aug 3 '18 at 20:35
  • $\begingroup$ What happens to $$-\frac{x\sin(2/x)}{4}?$$ $\endgroup$ – Omojola Micheal Aug 4 '18 at 13:09

You are on right track. And the challenge is to evaluate $$\lim_{x\to 0}\frac{1}{x}\int_{0}^{x}\cos(2/t)\,dt\tag{1}$$ Let $f(x) =\cos(2/x),x\neq 0,f(0)=0$ and then the limit above can be written as $F'(0)$ where $F(x) =\int_{0}^{x}f(t)\,dt$. Next consider the derivative evaluation $$\frac{d} {dx} (x^2\sin(2/x))=2x\sin(2/x)-2\cos(2/x)\tag{2}$$ so that if $G(x) =x^2\sin(2/x),x\neq 0, G(0)=0$ and $g(x) =2x\sin(2/x),x\neq 0,g(0)=0$ then equation $(2)$ can be written as $$G'(x) =g(x) - 2f(x),\forall x\in\mathbb {R} \tag{3}$$ (verify that the above holds for $x=0$ also). Since $f, g$ are Riemann integrable it follows that $G'$ is also Riemann integrable and $$G(x)=G(x) - G(0)=\int_{0}^{x}G'(t)\,dt\\=\int_{0}^{x}g(t)\,dt-2\int_{0}^{x}f(t)\,dt=\int_{0}^{x}g(t)\,dt-2F(x)\tag{4}$$ It follows by differentiating the above equation that $$G'(x) =g(x) - 2F'(x)$$ (note that $g$ is continuous everywhere) and hence $F'(0)=(g(0)-G'(0))/2=0$.


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