lim(n → ∞) π/n [sin π/n + sin 2π/n +......+ sin (n-1)π/n ] is equal to - Sarthaks eConnect | Largest Online Education Community
lim)┬(n→∞)〖(sin π/2n.sin 2π/2n.sin 3π/2n…….sin ((n-1)π)/2n)^(1/n) 〗 is equal to: a)1/2 b) 1/3 - YouTube
If I(n)=int(0)^(pi/2) sin^(x)x dx, then show that I(n)=((n-1)n)I(n-2). Hence prove that I(n)={(((n-1)/n)((n-3)/(n-2))((n-5)/(n-4))………(1/2)(pi)/2,"if",n"is even"),(((n-1)/n)((n-3)/(n-2))((n-5)/(n-4))………(2/3)1,"if",n"is odd"):}
The value of limit n∞ 1n^2 { sin^3 pi4n + 2sin^3 2pi4n + ... + nsin^3 npi4n } is equal to
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![Calculus] Can anyone explain in layman's terms why the limit of sin(pi/x) as x --> 0 is null? : r/learnmath](https://external-preview.redd.it/qbxOoh3XQjKwU39BU5kpfv0ug4Z0NVdfURsusIjhzms.jpg?auto=webp&s=af6c0c060d09dc25f75cecdd097001cba2e7d6cd "Calculus] Can anyone explain in layman's terms why the limit of sin(pi/x) as x --> 0 is null? : r/learnmath")
Calculus] Can anyone explain in layman's terms why the limit of sin(pi/x) as x --> 0 is null? : r/learnmath
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![IIT 2004] Find the limit of (2/PI)*(n+1)*arccos(1/n) - n as n tends to infinity. - YouTube](https://i.ytimg.com/vi/akawGXHiV3c/maxresdefault.jpg "IIT 2004] Find the limit of (2/PI)*(n+1)*arccos(1/n) - n as n tends to infinity. - YouTube")
IIT 2004] Find the limit of (2/PI)*(n+1)*arccos(1/n) - n as n tends to infinity. - YouTube
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The value of limit n→∞ n^ksin^2(n!)n + 2,0 < k < 1 , is equal to
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![limit n→∞pi/n [ sinpi/n+sin2pi/n+ .... +sin (n - 1 )pi/n ] is equal to :](https://haygot.s3.amazonaws.com/questions/1954614_1038211_ans_1fb8b3c433424f49823729fa116112f1.jpeg "limit n→∞pi/n [ sinpi/n+sin2pi/n+ .... +sin (n - 1 )pi/n ] is equal to :")
limit n→∞pi/n [ sinpi/n+sin2pi/n+ .... +sin (n - 1 )pi/n ] is equal to :
sin(3pi/2 + x) | sin(3pi/2 + theta) - YouTube
