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I have to calculate $$\lim_{x\to 0} {1\over x} \int_0^x \cos(t^2)\,dt$$

My intuition is that the answer is 1 because as $x$ becomes very small, $x^2$ also becomes very small and I am tempted to write $$\lim_{x\to 0} \int_0^x \cos(t^2)\,dt=\lim_{x\to 0} \int_0^x \cos(t)\,dt$$

And then, because $(\sin x)'=\cos x$, we have $$\lim_{x\to 0} {1\over x} \int_0^x \cos(t^2) \, dt = \lim_{x\to 0} {1\over x} \int_0^x \cos(t) \, dt=\lim_{x\to 0} {1\over x} (\sin x -\sin 0)=\lim_{x\to 0} {\sin x\over x}= 1$$ (well known limit solved with L'Hôpital's rule)

But I am pretty sure this is not rigorous and I am doing something I am not 'allowed to', especially the first equality I wrote.

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Define $F(x) = \int_0^x\cos(t^2)\,dt$. By the fundamental theorem of calculus, $F'(x) = \cos(x^2)$ and so $$ \lim_{x\to 0}\frac 1x\int_0^x\cos(t^2)\,dt = \lim_{x\to 0}\frac{F(x)-F(0)}{x-0} = F'(0) =1. $$

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Or go through by Integral Mean Value:

\begin{align*} \dfrac{1}{x}\int_{0}^{x}\cos(t^{2})dt=\cos(\eta_{x}^{2})\rightarrow\cos 0=1, \end{align*} where $\eta_{x}$ is in between $0$ and $x$.

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The fundamental theorem of calculus tells you that $$ \frac d {dx} \int_0^x \cos(t^2)\, dt = \cos(x^2).$$

The usual definition of derivative tells you that $$ \lim_{x\to0} \frac{\int_0^x \cos(t^2) \, dt - \int_0^0 \cos(t^2)\, dt}{x-0} = \left.\frac d {dx} \int_0^x \cos(t^2) \, dt \right|_{x\,=\,0}.$$

So you get $\cos(0^2)$ (which is $1$).

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  • $\begingroup$ Michael.A typo second last line upper integration limit should be x, not 2. $\endgroup$ Oct 28, 2019 at 19:23
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    $\begingroup$ Good to see you here again :) $\endgroup$
    – Botond
    Oct 29, 2019 at 0:41
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I'd say you're doing too much work; by L'Hopital's rule and the fundamental theorem of calculus,

$$\lim_{x\to0}\frac{\int_0^x\cos(t^2)\,\mathrm dt}x=\lim_{x\to0}\frac{\cos(x^2)}1=1$$

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    $\begingroup$ This works just fine, but the use of L'Hopital's rule is unnecessary when you're really just using the definition of the derivative. $\endgroup$ Oct 28, 2019 at 19:04
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Of course, this answer is overkill, but it does illustrate the power of Lebesgue's Density theorem, to take care of exercises like this one. Using that theorem and the fact that cos is even

\begin{align}\dfrac{1}{x}\int_{0}^{x}\cos(t^{2})dt=\dfrac{1}{2x}\int_{-x}^{x}\cos(t^{2})dt\to \cos 0=1.\end{align}

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  • $\begingroup$ So it is some sort of Lebesgue Differentiation Theorem, but you know it says only for a.e. whether the point $0$ is targeted is not known. $\endgroup$
    – user284331
    Oct 28, 2019 at 19:32
  • $\begingroup$ if $f$ is continuous it holds $everywhere.$ Inspect the proof to see that this is true. $\Big|\frac{1}{|B(x,r)|}\int_{B(x,r)}f(y)\;dy-f(x)\Big|\leq \frac{1}{B(x,r)}\int_{B(x,r)}|f(x)-f(y)|\;dy\leq\varepsilon$ $\endgroup$ Oct 28, 2019 at 20:20

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