Calculus Help!!!

Question

Evaluate lim x→0 (2sin(x)−sin(2x))/(2e^x −2−2x−x^2)

Answer 1

just plug in 0 for x, but if you get 0/0 that's undefined, indeterminant

then take derivatives separately of numerator and denominator

then plug in 0 again but you get 0/0 again

keep trying, more derivatives then plug in 0 until you get a real number

and that gives you the limit as x goes to zero = 3

(2sinx - sin2x)/(2e^x - 2 - 2x -x^2)

take derivatives: = (2cosx - 2cos2x)/(2e^x -2 -2x)

plug in x= 0 gives (2-2)/2-2) = 0/0

take derivatives again = (-2sinx +4sin2x)/(2e^x -2)

x= 0 gets you 0/0 again

take derivatives again = (-2cosx +8cos2x)/(2e^x)

plug in 0 to get (-2+8)/2 = 6/2 = 3

this method is called L'Hopital's rule or also spelled like Hospital

Answer 2

The first thing to do is to plug in 0 in for x and solve.

(2sin(0)−sin(2(0)))/(2e⁰ −2−2(0)−(0)²) = (0-0)/(2-2-0-0) = 0/0 This scenario of 0/0 allows you to apply L'Hospital's rule, take the derivative of the numerator and denominator.

The derivative of the numerator and denominator:

(2cos(x)-2cos(2x))/2e^x −2−2x) reduce by cancelling out a 2 from all terms

(cos(x)-cos(2x))/(e^x −1−x) plug in 0 in for x and solve

(cos(0)-cos(20))/(e¹ −1−0)

(1-1)/(1-1-0) This also results in 0/0, so we would apply L'Hospital's rule again, take the derivative of the numerator and denominator.

(-sin(x)+2sin(2x))/(e^x −1) plug in 0 in for x and solve

(0-0)/(1-1) This also results in 0/0, so we would apply L'Hospital's rule again, take the derivative of the numerator and denominator.

(-cos(x)+(2)(2)cos(2x))/(e^x) plug in 0 in for x and solve

(-cos(0)+2cos(2(0)))/(e)⁰)

(-1+4(1))/1 = 3/1 = 3