in Calculus edited by
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Let $\psi : \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function with $\psi(y) =0$ for all $y \notin [0,1]$ and $\int_{0}^{1} \psi(y) dy=1$. Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a twice differentiable function. Then the value of $$\lim _{n\rightarrow \infty}n \int_{0}^{100} f(x)\psi(nx)dx$$ is

  1. $f(0)$
  2. $f’(0)$
  3. $f’’(0)$
  4. $f(100)$
in Calculus edited by
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1 Answer

4 votes
4 votes
Best answer
Put $nx = t$ , $ndx = dt$ and $x = t/n$ we get
$\lim_{n \rightarrow \infty }n \int_{0}^{100} f(x) \psi (nx) dx = \lim_{n \rightarrow \infty }\int_{0}^{100n} f(t/n) \psi (t) dt$

$ = \int_{0}^{1} f(0) \psi (t)dt$ .

The last step is really important $t/n$ approaches 0 . $100n>>1$ and we dont have to go beyond 1 because $\psi (y) = 0$ elsewhere as given in question.

Continuing on $ = f(0)  \int_{0}^{1}  \psi (t)dt = f(0)* 1 = f(0)$ .

So $A$ is correct.
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4 Comments

I have seen the reason you are referring to, but it is not sufficient. Without the change of $\lim$ and $\int$ signs, you can’t claim those two quantities to be equal. Right now, I am unable to recall some examples for it, but certainly, there are many examples.
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Ok. I think you are objecting to the substitution $n dx = dt$. One of the underlying assumption in the whole analysis is $n$ and $x$ are independent.
Let me know of the examples you are talking about.
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I think Uniform continuity of f allow us to interchange lim and integration sign, so it is correct.
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