# Math

posted by .

Find the limit of
(2cosx+3cosx-2)/(2cosx-1)as x tends to pi/3.

• Math -

if you let y=cosx, then you have

(2u-1)(u+2)/(2u-1)

so, except when u=1/2, you just have (u+2)

so, the limit is cos pi/3 + 2 = 5/2

there is a removable hole at x=pi/3

## Similar Questions

1. ### Trig

prove the identity (sinX)^6 +(cosX)^6= 1 - 3(sinX)^2 (cosX)^2 sinX^6= sinx^2 ^3 = (1-cosX^2)^3 = (1-2CosX^2 + cos^4) (1-cosX^2) then multiply that out 1-2CosX^2 + cos^4 - cosX^2 + 2cos^4 -cos^6 add that on the left to the cos^6, and …
2. ### Math - Solving for Trig Equations

For the following equation, why cannot I not solve it?
3. ### math

tanx+secx=2cosx (sinx/cosx)+ (1/cosx)=2cosx (sinx+1)/cosx =2cosx multiplying both sides by cosx sinx + 1 =2cos^2x sinx+1 = 2(1-sin^2x) 2sin^2x + sinx-1=0 (2sinx+1)(sinx-1)=0 x=30 x=270 but if i plug 270 back into the original equation …
4. ### trig

prove that (2cosx+1)(2cosx-1)= 1 +2cos2x
5. ### RESPONSE to PreCalc Question

h t t p : / / w w w . j i s k h a . c o m / d i s p l a y . c g i ?
6. ### Calculus

Determine the concavity and find the points of inflection. y=2cosx + sin2x 0≤x≤2pi y'=-2sinx + 2cos2x y"=-2cosx-4sinx How do I find the IP(s)?
7. ### RSNHS

What is the limit of 1-2x^2 -2cosx + cos^2x all over x^2?
8. ### Pre Calculus

Having trouble verifying this identity... cscxtanx + secx= 2cosx This is what I've been trying (1/sinx)(sinx/cosx) + secx = 2cosx (1/cosx)+(1/cosx) = 2 cosx 2 secx = 2cosx that's what i ended up with, but i know it's now right. did …
9. ### derivatives

x^2cosx-2xsinx-2cosx
10. ### Trigonometry (identities) (last help needed)

1. (2cosx-2secx)(3cosx^2+3secx^2+β) 2. (16tanxsecx-4secx)/(8tanxsecx+4secx) 3.sinxcosx^3+cosx^4 (hint: GCF) 4. (cosβ+sinβ)^2

More Similar Questions