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Question-9057




Question Number 9057 by sandipkd@ last updated on 16/Nov/16
Answered by aydnmustafa1976 last updated on 16/Nov/16
nsin(1/n)=lim((sin(1/n))/(1/n))=lim((sint)/t)=1 therefore  4∫_0 ^1 (1/(x^2 +1))dx=4.arctgx∣_0 ^1 =4((Π/4)−0)=Π
nsin1n=limsin1n1n=limsintt=1therefore4011x2+1dx=4.arctgx01=4(Π40)=Π
Commented by sandipkd@ last updated on 17/Nov/16
thanks..i was confused in nsin 1/n
thanks..iwasconfusedinnsin1/n
Answered by aydnmustafa1976 last updated on 17/Nov/16
lim _(n→∞) ∫_0 ^(nsin(1/n)) ..... =∫_0 ^(lim(1/(1/n)).sin(1/n)) ....=∫_0 ^(lim((sin(1/n))/(1/n))) ...=....
limn0nsin1n..=0lim11n.sin1n.=0limsin1n1n=.
Answered by aydnmustafa1976 last updated on 17/Nov/16
note: lim_(n→∞) ((sin(1/n))/(1/n))=li_((1/n)→0) m((sin(1/n))/(1/n))=lim_(t→0) ((sint)/t)=0
note:limnsin1n1n=lim1n0sin1n1n=limt0sintt=0
Answered by aydnmustafa1976 last updated on 17/Nov/16
sorry lim((sint)/t)=1 when t→0
sorrylimsintt=1whent0

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