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Consider a sequence for functions $f_n: [0,2] \to \mathbb{R}$ such that $f_n(0) = 0$ and $f_n(x) = \frac{\sin (x^n)}{x^n}$ for $x \in (0, 2]$. Find $$\lim_{n \to \infty} \int_{[0,2]} f_n(x) \, dx$$

I'm not sure how to begin with one. Some ideas that come to mind are:

  • dominated/bounded convergence theorem (but it's not clear to me what function this sequence converges to)
  • using absolute continuity (if it can be shown) to utilize FTC

Any ideas on how to approach this one? Let me know any hints or suggestions. Thanks.

robjohn
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Grigor Hakobyan
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    Define $f(x)=\lim_{n\rightarrow\infty} f_n(x)$. So, if $x\in(0,1)$, $f(x)=\lim_{x\rightarrow 0} \frac{\sin(x)}{x}=1$. If $x=1$, we have that $f(x)=f(1)=\sin(1)$. If $x\in(1,2]$, we have that $f(x)=\lim_{x\rightarrow\infty} \frac{\sin(x)}{x}=0$. So $f(x)=\begin{cases} 1&\text{if}, x\in (0,1)\ \sin(1)&\text{if}, x=1\ 0&\text{if}, x\in (1,2] \end{cases}$ – Tio Zuca Dec 15 '23 at 17:17
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    From this, we have that $\lim_{n\rightarrow\infty}\int_0^2 f_n(x)dx=\int_0^2 f(x)dx = \int_0^1 f(x)dx+\int_1^2 f(x)dx=1+0$ – Tio Zuca Dec 15 '23 at 17:21
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    In fact, we would need to prove that the limit definition of $f(x)$ converges to the piecewise function, and use the (maybe) fact that $\lim_{n\rightarrow\infty}\int_a^b f_n(x)dx=\int_a^b \lim_{n\rightarrow\infty} f_n(x) dx$ – Tio Zuca Dec 15 '23 at 17:25
  • @TioZuca Thanks for all the info. What allows us to assert $\lim_n \int f_n = \int \lim_n f_n$? Can we show $f_n \nearrow f$? – Grigor Hakobyan Dec 15 '23 at 17:30
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    @GrigorHakobyan $f_n$ are bounded by $1$ – Kroki Dec 15 '23 at 17:33
  • Or many similar posts – Anne Bauval Dec 15 '23 at 17:51

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