is it always satisfying?
A=lim[n→∞]∫f(n,x)dx
B=∫lim[n→∞]f(n,x)dx
A=B??
please show counter example
checking
(1)
f(n,x)=x^n ,x[0→1]
A=lim_(n→∞) ∫_0 ^1 x^n dx=lim_(n→∞) (1/(n+1))x^(n+1) =0
B=∫_0 ^1 lim_(n→∞) x^n dx=∫_0 ^1 0dx=0
so A=B
(2)
f(n,x)=(1+(x/n))^n
A=lim_(n→∞) ∫(1+(x/n))^n dx
=lim_(n→∞) (n/((n+1)))(1+(x/n))^(n+1)
=e^x
B=∫lim_(n→∞) (1+(x/n))^n dx
=∫e^x dx
=e^x
so A=B
. . .
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