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: $$\int{\biggl(\lim_{x\to n}\dfrac{nx}{x^n}\biggl)}dx$$
Limit inside an integral
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$\begingroup$
calculus
integration
limits
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0Limit will return you a constant , and integrating a constant will give you a linear equation. – 2012-04-27
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1Your 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. – 2012-04-27
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0In general, having a limit inside your integral is no big deal. What you have to be careful about, is having a *limit of integrals* – 2012-04-27
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0$\lim\limits_{x\to n}{\dfrac{nx}{x^n}}$ is a function of $n$. $x$ is a bound variable in the expression. – 2012-04-27
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0i 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 – 2012-04-27
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0we will calculate the integral of a constant. – 2012-04-27
2 Answers
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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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1the limit is not equal to $n$ – 2012-04-27
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0sorry , a silly mistake! – 2012-04-27
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0we 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. – 2012-04-27
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0@AbdelmajidKhadari: You're missing an $x$ after integrating. – 2012-04-27
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0i miss an $x$, where !? i think we have $\int dx=c$. – 2012-04-27
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0No, $\int \ \mathrm{d}x = x + C$. – 2012-04-27
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0@SauravTomar What's $L$ in this equation? $L=n^{2-n}$ – 2012-04-27
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0@Garmen1778 yes L(limit) is $n^{2-n}$ – 2012-04-28