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!

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

I can't use that rule yet.

I have to use limit only to find the answer.

sin 2 x ---- 2 x + .... as x -->0

tan x also ---> x
so
2x /(2x - x)

2x/x = 2

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

To find the limit of the given function as x approaches 0, we can simplify and apply some algebraic steps. Here's how you can do it:

Step 1: Start with the given function:
lim(x->0) [(sin(2x))/(2x - tan(x))]

Step 2: Use the trigonometric identity: sin(2x) = 2sin(x)cos(x)
lim(x->0) [(2sin(x)cos(x))/(2x - tan(x))]

Step 3: Factor out a sin(x) from the numerator:
lim(x->0) [sin(x)(2cos(x))/(2x - tan(x))]

Step 4: Simplify the denominator using the trigonometric identity: tan(x) = sin(x)/cos(x)
lim(x->0) [sin(x)(2cos(x))/(2x - sin(x)/cos(x))]

Step 5: Combine the fractions in the denominator:
lim(x->0) [sin(x)(2cos(x))/(2x(cos(x)) - sin(x))]

Step 6: Multiply both the numerator and denominator by cos(x):
lim(x->0) [sin(x)(2cos(x))(cos(x))/(2x(cos(x))(cos(x)) - sin(x)(cos(x))]

Step 7: Simplify the denominator:
lim(x->0) [2sin(x)cos^2(x)]/[2xcos^2(x) - sin(x)cos(x)]

Step 8: Divide both the numerator and denominator by x:
lim(x->0) [2sin(x)cos^2(x)]/[xcos^2(x) - sin(x)cos(x)]

Step 9: Substitute x = 0 into the expression:
[(2sin(0)cos^2(0))]/[(0)(cos^2(0)) - sin(0)cos(0)]

Step 10: Simplify using trigonometric identities:
[0]/[0 - 0] = 0/0

At this point, we have an indeterminate form of 0/0, which means we need to perform additional steps to evaluate the limit.

Step 11: Use L'Hopital's Rule: Differentiate both the numerator and denominator with respect to x:
lim(x->0) [(2cos^2(x) - 2sin^2(x))(1)]/[-(cos^2(x) + 2sin(x)cos(x))]

Step 12: Simplify the numerator:
lim(x->0) [2(cos^2(x) - sin^2(x))]/[-(cos^2(x) + 2sin(x)cos(x))]

Step 13: Apply trigonometric identities to the numerator:
lim(x->0) [2(1 - sin^2(x))]/[-(cos^2(x) + 2sin(x)cos(x))]

Step 14: Simplify the numerator:
lim(x->0) [2cos^2(x)]/[-(cos^2(x) + 2sin(x)cos(x))]

Step 15: Substitute x = 0 into the expression:
[2cos^2(0)]/[-(cos^2(0) + 2sin(0)cos(0))]

Step 16: Evaluate the trigonometric values:
[2(1)]/[-(1 + 0)] = -2/1

Final Step: Simplify the result:
-2

Therefore, the limit of (sin(2x))/(2x - tan(x)) as x approaches 0 is -2.