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If

$$\lim_{x\to0}\frac1{x^m}\prod_{k=1}^n \int_0^x\big[k-\cos(kt)\big]\mathrm dt$$

exists and is equal to $20$ (where $m,n\in\mathbb N$) then what is the value of $n$?

I started this question with the general term and am slightly confused on whether to apply L'hospital method on this particular problem, are there any other ways to evaluate this?

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  • $\begingroup$ Possible hint: logarithms. $\endgroup$ Commented Mar 20, 2023 at 5:16
  • $\begingroup$ Use the fundamental theorem of calculus to evaluate limit of $\frac{1}{x}\int_0^x (k-\cos kt) \, dt$ and then you can easily figure out rest of the solution. $\endgroup$
    – Paramanand Singh
    Commented Mar 20, 2023 at 13:38

2 Answers 2

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$\textbf{Hint:}$ For $n>1$, we have that $n - \cos n t \sim n-1$ as $t\sim 0$, so each term beyond the first one only needs to absorb one factor of $x$ in the denominator in order to have a nonzero constant limit, i.e. heuristically

$$\frac{\displaystyle{\int_0^x} \left(n - \cos nt \right)\:dt}{x} \sim \frac{(n-1)\cdot x}{x} = n-1$$

For the first term $n=1$, use the identity

$$\cos 2\theta = 1 - 2\sin^2\theta$$

and the limit

$$\lim_{z\to0}\frac{\sin z}{z} = 1$$

to figure out how many $x$'s need to be absorbed into that first term to have a nonzero constant limit. What does this tell you about the relationship between $m$ and $n$?

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    $\begingroup$ Or L'Hopital's rule will show more exactly that $\lim_{x \to 0} \frac 1x \int_0^x (k-\cos kt)\, dt = \lim_{x \to 0} (k-\cos kx) = k-1$. $\endgroup$
    – aschepler
    Commented Mar 20, 2023 at 13:25
  • $\begingroup$ @aschepler absolutely! But I feel L'Hopital misses the point of when it is necessary. Here, the intuition built is the fundamental theorem of calculus that the slope of the antiderivative function is just the value of the integrand at that point, just as one of the comments suggests. The intuition lost from just hammering away at L'Hopital's rule is of how to distribute the $x$'s and when extra are necessary (i.e. when the integrand has a critical zero), which is reduced to arbitrary guess and check by LH. $\endgroup$ Commented Mar 20, 2023 at 17:21
  • $\begingroup$ Your argument is certainly why one would attempt to find the limit with factor $\frac 1x$ in the first place, yes. LH is just a nice shortcut in proving the limit formally, as opposed to bounding and squeezing or whatever other techniques. $\endgroup$
    – aschepler
    Commented Mar 20, 2023 at 17:53
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First of all, it results that

$\begin{align}\int_0^x\big[k-\cos(kt)\big]\mathrm dt&=\left[kt-\frac{\sin(kt)}k\right]_0^x=kx-\frac{\sin(kx)}k=\\[2pt]&=x\left[k-\frac{\sin(kx)}{kx}\right]\end{align}$

for any $\;k\in\Bbb N\;\land\;1\leqslant k\leqslant n\,.$

Consequently,

$\displaystyle\frac1{x^m}\prod_{k=1}^n \int_0^x\big[k-\cos(kt)\big]\mathrm dt=\frac1{x^{m-n}}\prod_{k=1}^n\left[k-\frac{\sin(kx)}{kx}\right]=$

$=\displaystyle\frac1{x^{m-n-2}}\!\cdot\!\frac{1-\frac{\sin x}x}{x^2}\!\cdot\!\prod_{k=2}^n\left[k-\frac{\sin(kx)}{kx}\right]=$

$=\displaystyle\frac1{x^{m-n-2}}\!\cdot\!\frac{x-\sin x}{x^3}\!\cdot\!\prod_{k=2}^n\left[k-\frac{\sin(kx)}{kx}\right].$

Given that $\;\lim\limits_{x\to0}\dfrac{x-\sin x}{x^3}=\dfrac16\;$ and $\;\lim\limits_{x\to0}\dfrac{\sin{(kx)}}{kx}=1\\$$\text{for any }\;2\leqslant k\leqslant n\;,\;\text{ it results that}$

$\displaystyle\lim\limits_{x\to0}\frac1{x^m}\!\prod_{k=1}^n\!\int_0^x\!\!\big[k-\cos(kt)\big]\mathrm dt=\begin{cases}\displaystyle\frac16\prod_{k=2}^n(k-1)&\text{if }\;m=n+2\\[5pt]\quad0&\text{if }\;m<n+2\\[5pt]\text{does not exist}&\text{if }\;m>n+2\end{cases}\\$

Since the limit exists and is equal to $\,20\,,\;$ it follows that $\;m=n+2\;$ and $\;\displaystyle\frac16\prod_{k=2}^n(k-1)=20\;,\;$ hence,

$\displaystyle\prod_{k=2}^n(k-1)=120\;\;,$

$(n-1)!=120=5!\;\;,$

$n-1=5\;\;,$

$n=6\,.$

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