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

[ilhamGD]

Homework Statement

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

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
Can u please help ?
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[HallsofIvy]

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?

 

[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)?

 

[haruspex]

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)?

I don't get the same answer for that limit. Please post your detailed working.