Find the limit of

(2cosx+3cosx-2)/(2cosx-1)as x tends to pi/3.

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

To find the limit of a function as x approaches a certain value, we can substitute that value into the function and evaluate. In this case, we need to find the limit of (2cosx+3cosx-2)/(2cosx-1) as x tends to π/3.

Let's substitute π/3 into the function:
(2cos(π/3) + 3cos(π/3) - 2)/(2cos(π/3) - 1)

To evaluate cos(π/3), we can use the unit circle or knowledge of trigonometric values. The cosine of π/3 is 1/2.

Substituting 1/2 for cos(π/3) in the expression:
(2 * (1/2) + 3 * (1/2) - 2)/(2 * (1/2) - 1)

Simplifying the expression:
(1 + 3 - 2)/(1 - 1)

Now we can calculate the value:
(2)/(0)

Notice that the denominator is equal to 0. When the denominator of a fraction is 0, the fraction is said to be undefined. In mathematics, we cannot divide by 0. Therefore, the limit does not exist in this case.

Hence, the limit of (2cosx+3cosx-2)/(2cosx-1) as x tends to π/3 does not exist.