# Find the value of a and b for which the function is continuous at 2

1. Sep 23, 2014

### ilhamGD

1. The problem statement, all variables and given/known data

$$u(x) = \begin{cases} \frac{3x+b}{4} & \text{if } x \geq 2 \\ \frac{(3-x)^n-a}{x-2} & \text{if } x < 2 \end{cases}$$
find the value of a and b for which the function is continuous at 2

3. The attempt at a solution
I tried to proof that lim(3x+b)/4 = lim (3-x)^n-a/x-2 = f(2)
that gives lim (3-x)^n-a/x-2= 6+b/4
But I have a problem with the limit when x< 2, I don't know how to solve it

2. Sep 23, 2014

### HallsofIvy

Staff Emeritus
The difficulty with x< 2 is, of course, that the denominator is 0 at x= 2. In order for that to have a chance at having a limit, the numerator must also be 0 at x= 2. That is, we must have $$(3- 2)^n- a$$. What does that tell you about a?

3. Sep 23, 2014

### ilhamGD

It means that a=1, and after using the formula for A^n-B^n the limit when x< 2 becomes -(2^n-1+2^n-2+....+1) which is -2^(n-2) and then I can find b which is -2^(n-1)/3
can u tell me if that is correct ? and should I also find f(2)?

4. Sep 23, 2014