# Piecewise function question

1. Oct 5, 2011

### BifSlamkovich

1. The problem statement, all variables and given/known data
With reference to this graph:
http://imageshack.us/photo/my-images/41/deletemev.jpg/

Is the value of c that makes lim x->1 [f(x) + c g(x)] exist equal to the same value of c that makes f(x) + c g(x) continuous at x = 1?

2. Relevant equations

3. The attempt at a solution

They're the same, right, because the value of c that makes the lim x->1 of f(x) + cg(x) exist is the same as the value of c that makes f(x) + cg(x) continuous at x=1? Or could I be wrong??

2. Oct 5, 2011

### HallsofIvy

Staff Emeritus
Yes, it is a basic theorem of limits that $\lim_{x\to a}(f(x)+ cg(x))= \lim_{x\to a}f(x)+ c\left(\lim_{x\to a} g(x)\right)$.

3. Oct 5, 2011

### BifSlamkovich

I'm referring to the continuity of a certain function, not the applicability of limits to an expression.

4. Oct 5, 2011

### SammyS

Staff Emeritus
Look again. What is f(1) + c g(1) ? I'm not saying it's Yes, or No, just that it doesn't necessarily follow.

5. Oct 5, 2011

### BifSlamkovich

f(1) + cg(1) from the left side has to equal f(1) + cg(1) from the right side in order for the lim x-> 1 to exist, i.e., an appropriate value of c has to be determined. But idk if it's the very same value that makes the function, f(1) + cg(1), continuous at x=1, because continuity has 3 criteria:
1. the function is defined at x=1, which I think it is
2. lim x-> 1 of f(1)+cg(1) exists, which I think it does because the piecewise discontinuities are filled and unfilled
3.lim x-> 1 f(1) + cg(1) = f(1) + cg(1), which seems to be the case because there is no removable discontinuity.

So I would be inclined to think that the function f(x) + cg(x) is indeed continuous at x=1, but I could be wrong.

6. Oct 5, 2011

### SammyS

Staff Emeritus
It looks to me as if $\lim_{x\,\to\,1}(f(x) + c\,g(x))=2$ if c = 2. Otherwise, f(x) + cg(x) is not continuous.

However, f(1) = 4, and g(1) = 0, so no matter what value you use for c, f(1) + cg(1) = 4.