Calculus Help Please!!!

show all of the algebraic steps in finding lim x->0 (sin2x)/(2x-tanx)

I know the answer is 2 but I need the steps thank you!

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  1. The key here is l'hospital's rule

    lim f/g = lim f'/g'

    so, the limit is the same as

    2cos2x / (2 - sec^2 x) = 2*1/(2-1) = 2

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  2. I can't use that rule yet.

    I have to use limit only to find the answer.

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  3. sin 2 x ---- 2 x + .... as x -->0
    tan x also ---> x
    so
    2x /(2x - x)

    2x/x = 2

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  4. hmmm. well, I guess you can't use Taylor series either, eh?

    (sin2x)/(2x-tanx)
    I assume you've shown that lim (sinx)/x = 1, so if our limit is L,

    1/L = (2x-tanx)/(sin2x)
    = (2x)/sin(2x) - tanx/sin2x
    = 1 - (sinx/cosx)/(2sinx cosx)
    = 1 - (1/cosx)/(2cosx)
    limit as x->0 is
    = 1 - 1/2
    = 1/2

    So, L = 2

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