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All about sequence of functions 
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#1
Oct505, 01:47 PM

P: 91

Let {h_n} be a sequence of function defined on the interval (0,1) where
h_n(x) = (n+n)x^(n1)(1x) a. find lim (n> +oo) (integral) (from 0 to 1) h_n(x) dx. b. show that lim (n> +oo) h_n(x) = 0 on (0, 1) c. Show that lim (n> +oo) (integral) (from 0 to 1) h_n(x) dx is not equal to integral (from 0 to 1) (0 dx). What went wrong? SOlutions: a. lim (n> +oo) integral (from 0 to 1) h_n(x) dx = lim (n> +oo) integral (from 0 to 1) (n+n)x^(n1)(1x) dx =lim (n> +oo)(n+n) integral (from 0 to 1)x^(n1)(1x) dx = lim (n> +oo)(n+n) (1/n  1/(n+1)) = lim (n> +oo)n(n + 1) (1/((n)(n+1)) = 1. b. I used the nth term test in proving this... because if the series of h_n(x) is convergent then lim (n> +oo) h_n(x) = 0 on (0, 1). But by ratio test, h_n(x) is convergent because the limit of a(n+1)/a(n) as n > +oo is x, but 0 < x < 1. c. That's the part that I got stuck... well, it seems that the statement above is true... how do I solve this? 


#2
Oct505, 03:14 PM

HW Helper
P: 2,566

Roughly speaking, the main contribution to the integral for large n comes from an increasingly small neighborhood near 1. For any given point x, no matter how close to one, h_{n}(x) eventually gets very small as n increases. But there are still points left between x and 1 for which the value of h_{n} evaluated at these points is very large, and this keeps the value of the integral at one. By the way, did you mean n+n^2 for the factor in front of the function?



#3
Oct505, 08:58 PM

P: 91

Thank you for clarifying...
yes, it should have been n^2 + n. 


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