Limit tanx-sinx/x^3 as x approaches 0 (without L'Hopital's rule)

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lim x-0 (tanx-sinx)/x^3 =1/2
Find the limit of tanx-sinx/x3 as x approaches 0 without using the L'Hospital's (L'Hopital's) rule.

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Why the limit is not equal to 0 since tanx/x = 1 and sinx/x = 1 then 1 - 1 = 0?

ahsingtv
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Why can we not write it as tanx/x^3 - sinx/x^3 and since sin or tan by x when tending to zero is 1, that will give 1÷x^2 - 1÷x^2 which is zero?

justinwhy
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Can I do (1-cosx)/x^2 times cosx = (1-cosx/x^2) times (1/cosx) then when x—>0 it’s 1/2 times 1 = 1/2 .
Using (1-cosx)/x^2= 1/2 when x—>0

frangou
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1-cox/x²cox
We can seperate
1-cox/x . 1/xcosx
(As identity says 1-cosx/x=0)
Ans is zero

trendypopcorn
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Oh for the sake of all the divine beings that may exist in this life, THANK YOU

bernarditamedina
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thank u very much sir, , u explained it very welll

Zeus-zhzx
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Why can't we use sinx/x x tends to 0 = 1 and Tanx/x x tends to 0 = 1

bhaavarora