Evaluate each limit

Question

lim_θ→0 (5cos2θ-5)/(4θ)

Answer 1

If you are unfamiliar with L'Hopital's Rule, we can use the following limit proved in Calculus classes or found in Calculus textbooks. 

 

    lim_x→0 [(cosx - 1)/x] = 0

 

The given limit is equivalent to (5/2)lim_θ→0 [ (cos2θ - 1) / (2θ) ]

 

    So, if we let u = 2θ, then as θ→0, u→0.

 

We have (5/2)lim_u→0 [(cosu - 1) / u)] = (5/2)(0) = 0

 

Answer 2

Use L'Hopital's rule.  Are you familiar with it?

 

lim_x→0 [5cos(2x) - 5]/(4x) = lim_x→0 [-10sin(2x)]/4 = 0