# 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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