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Well, I mean, imagine that you have a function: $$f(x)=\lim\limits_{x\to n}{\dfrac{nx}{x^n}}$$ Would it be possible to write an integral of that? Something like this: $$\int{\biggl(\lim_{x\to n}\dfrac{nx}{x^n}\biggl)}dx$$

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Limit will return you a constant , and integrating a constant will give you a linear equation. –  Tomarinator Apr 27 '12 at 18:39
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Your first bit doesn't make sense as a function of $x$ since $x$ is a dummy variable. As a function of $n$, however, it is fine. –  J. M. Apr 27 '12 at 18:40
    
In general, having a limit inside your integral is no big deal. What you have to be careful about, is having a limit of integrals –  M Turgeon Apr 27 '12 at 18:46
    
$\lim\limits_{x\to n}{\dfrac{nx}{x^n}}$ is a function of $n$. $x$ is a bound variable in the expression. –  Thomas Andrews Apr 27 '12 at 19:03
    
i think that you mean $n\to +\infty$ because for the case $x\to n$ the expresion of $\lim_{x\to n}\dfrac{nx}{x^n}$ will not depend on $x$ will just depend on $n$ is not interesting –  Abdelmajid Khadari Apr 27 '12 at 19:16

2 Answers 2

up vote 2 down vote accepted

yes, you can do it but if $\lim\limits_{n\to +\infty}f_{n}(x)=f(x)\in L^1$

$$\int \lim_{n\to +\infty}f_{n}(x)dx=\int f(x)dx$$

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Of course you can do it,

$$\displaystyle\int{\biggl(\lim\limits_{x\to n}\dfrac{nx}{x^n}\biggl)}dx= \displaystyle\int L.dx = Lx +c $$

$L= n^{2-n}$

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the limit is not equal to $n$ –  Beni Bogosel Apr 27 '12 at 18:42
    
sorry , a silly mistake! –  Tomarinator Apr 27 '12 at 18:45
    
we have $\displaystyle\int L.dx=\displaystyle\int n^{2-n}.dx=n^{2-n}\displaystyle\int dx= n^{2-n}.c$, wher $c$ is a constant. –  Abdelmajid Khadari Apr 27 '12 at 19:32
    
@AbdelmajidKhadari: You're missing an $x$ after integrating. –  Javier Badia Apr 27 '12 at 19:56
    
i miss an $x$, where !? i think we have $\int dx=c$. –  Abdelmajid Khadari Apr 27 '12 at 20:02

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