# Homework Help: Continuity of the function

1. Nov 13, 2008

### l46kok

1. The problem statement, all variables and given/known data
$$f(x) = x^2 - 4x + a$$
$$g(x) = \lim_{n\rightarrow\infty} \frac {2|x-b|^n + 1}{|x-b|^n + 1}$$

let $$h(x) = f(x)g(x)$$

Find the sum of a+b that makes h(x) continuous for all x.

2. Relevant equations
Power Series??? Derivation to test continuity???

3. The attempt at a solution

Well I know a function is continuous if

1)f(c) exists
2)Limit as x goes to c of f(x) exists
3)#1 is equal to #2

But I don't know where to go from here. Please help.

2. Nov 13, 2008

### Office_Shredder

Staff Emeritus
How does it fail to be continuous? f(x) is continuous for all a, but g(x) has a jump from when |x-b| < 1 (in which case g(x)=1) to when |x-b|>1 (in which case g(x)=2). So you need to figure out how to take care of it.

3. Nov 13, 2008

### l46kok

I understood why the discontinuity occurs, since |x-b|^n of n to the infinity could take you to 0 or a huge value (I think it was related to geometrical sum) depending on whether it's smaller than 1 or bigger than 1. I still do not have a clue how to get the specific values of a and b from it...

Last edited: Nov 13, 2008
4. Nov 13, 2008

### Office_Shredder

Staff Emeritus
So, for f(x)g(x) to be continuous, you know that when |x-b| isn't near 1, the function's continuous. You should realize this gives us that the only two points of concern are x=b+1 and x=b-1. You have that as x approaches b+1 and b-1 from different sides, g is either 2 or 1. So you need f(b+1) and f(b-1) to be values such that f(b-1)*2=f(b-1)*1 and f(b+1)*2=f(b+1)

5. Nov 13, 2008

### l46kok

Then that simply means

f(b-1) = 1
f(b+1) = 1

So we have

$$1 = (b-1)^2 - 4(b-1) + a$$
$$1 = (b+1)^2 - 4(b+1) + a$$

$$0 = (b-1)^2 - 4(b-1) - (b+1)^2 + 4(b+1)$$
$$0 = (b-1)^2 - (b+1)^2 + 8$$

$$b = 2$$

Then,

$$1 = (2+1)^2 - 4(2+1) + a$$
$$1 = 9 - 12+ a$$
$$a = 4$$

a + b = 6 is what I get, but supposedly this isn't the answer. I don't see what went wrong here, everything seems right.

Actually, I thought about it, but for the two distinct points of x = b-1 and x = b+1 to be continuous, shouldn't that mean that
f(b-1) = f(b+1) ?

6. Nov 13, 2008

### Office_Shredder

Staff Emeritus
f(b-1)=1. Then 2*f(b-1)=f(b-1) gives us 2=1. Are you sure that's the right value?

f(b-1) and f(b+1) have nothing to do with each other really (although it is correct that f(b-1)=f(b+1) is necessary, this is more by coincidence than any deep connection).

7. Nov 13, 2008

### l46kok

Oh, it's actually

f(b-1) = 0
f(b+1) = 0

Doh! Thx

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