Evaluate the limit of expression 4cosx-6sin^2x+3tan^2x. x-->x0 , x0=60 degrees
2 Answers
justaguide | College Teacher | (Level 2) Distinguished Educator
We have to find the value of lim x--> 60 degrees[ 4cos x - 6(sin x)^2 + 3*tan x)^2]
Here we see that cos 60 , (sin 60)^2 and (tan 60)^2 are defined, so we can just substitute x with 60.
Substituting x = 60 degrees
4cos x - 6(sin x)^2 + 3*tan x)^2
=> 2 - 6*(3/4) + 3* 3
=> 2 - 4.5 + 9
=> 6.5
The required value of lim x--> 60 degrees[ 4cosx - 6sin^2x + 3tan^2x] is 6.5
giorgiana1976 | College Teacher | (Level 3) Valedictorian
To evaluate the limit, we'll have to substitute x by the value of x0, in the given expression.
lim (4cosx-6sin^2x+3tan^2x) = 4cos 60 - 6*(sin 60)^2 + 3*(tan 60)^2
cos 60 = 1/2
sin 60 = sqrt3/2
We'll raise to square both sides:
(sin 60)^2 = (sqrt3/2)^2
(sin 60)^2 = 3/4
tan 60 = sqrt 3
We'll raise to square both sides:
(tan 60)^2 = 3
We'll substitute the values of the functions in the expresison above:
lim (4cosx-6sin^2x+3tan^2x) = 4(1/2) - 6*(3/4) + 3*(3)
lim (4cosx-6sin^2x+3tan^2x) = 2 - 9/2 + 9
lim (4cosx-6sin^2x+3tan^2x) = (4 - 9 + 18)/2
lim (4cosx-6sin^2x+3tan^2x) = 13/2, for x-> 60 degrees