Calculus Help!!!

Question

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

Answer 1

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