What is limit tend to 0 Of ( cos x - 1 ) / ( 2x^2 )

Question

( cos x - 1 ) : 2x^2

Answer 1

Mohamad,

 

Apply L'Hospitals rule twice to get rid of the indeterminate division by zero.

 

lim_x→0 [cos(x) - 1]/(2x²) = lim_x→0 [-sin(x)]/(4x) = lim_x→0 [-cos(x)]/4 = -¼

Answer 2

if you apply L'Hopitals rule, you take the derivative of each function and evaluate the limit

 

(cosx - 1)/2x²

 

-sinx / 4x     take the derivative again

 

-cosx / 4

 

now apply the limit x->0

 

-1/4