Ayman S.
asked 02/17/19Any tips to solve these limits ?
Concidering that f is continious over the interval [0;1]
lim n→∞ ∫01 ( f (x) /x+n ) dx
lim n→∞ ∫01 (xn f (x) ) dx
lim n→∞ ∫01 (f (x)/1+nx) dx
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lim n→∞ ∫0n (xn e-nx ) dx
1 Expert Answer
I am not sure I can solve all of these for you, but I can give you one big hint:
since f is continuous on the interval, it is bounded. Call the bound B.
In the first one the integral is less than B*∫dx/(x+n) = B*ln[(n+1)/n] = 0.in the limit
In the 2nd one the integral is less than B∫xn dx = B*[1/n+1] = 0 in the limit
In the 3rd one the integral is less tan (B/n)*∫dx/{x+(1/n)] = (B/n)ln](n+1)/n] = 0 in the limit
I will have to think about the 4th one & if I can give you further help, I will get back to you.
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Ayman S.
for the first too limits we can use the fact that there is m,M from IR / m = min (f(x) ) over [0,1] and M= max(f(x)) over [0,1] we will use that along with the "Squeeze Theorem" to find our limit but ... What about the others ?02/17/19