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lim x â†’ pi2 cos x³√ (1-sin x)²

#Mathematics #Calculus

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Hodge Conjecture ·2010-08-06 09:48:15

lim xâ†’0 x tan 2x - 2x tanx (1-cos2x)²

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AKHIL ·2010-08-06 10:10:56

u can solve second one by usin L hospital's rule...

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qwerty ·2010-08-06 23:14:43

for second one , convert tan into sin and cos , and simplify

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Avik ·2010-08-07 04:07:37

1) Substituting x= pi/2 +h where hâ†’0.

Lim(hâ†’0) cos(pi/2 +h)[1-sin(pi/2+h)]^2/3

= Lim(hâ†’0)sin(h)(1-cos(h))^2/3

Now u can use series expansion fr Sin(h) & cos(h) as hâ†’0.

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Hodge Conjecture ·2010-08-07 07:20:54

see i hv solved the 1st sum.... but the am getting the ans. of form non-zero0...implying limit does not exist.......
can anyone check this ans. , if its correct or not???

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Basic Integral Calculus Operators Symbols Matrix Cases

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° ∂ ∞ α β γ δ θ λ μ ν ξ π ρ σ φ ω ε ƒ κ τ υ Γ Δ Λ Σ Π Ω

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