Calculus

Start Your Free Trial

Evaluate the limit of expression 4cosx-6sin^2x+3tan^2x. x-->x0 , x0=60 degrees

| Certified 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

Approved by eNotes Editorial

giorgiana1976 | Student

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

Approved by eNotes Editorial

Already a member? Log in here.