# Find a value of the constant k such that the limit exists

1. Sep 17, 2006

### Jacobpm64

Find a value of the constant k such that the limit exists.

lim (x2 - kx + 4) / (x - 1)
x->1

We could do...
just try number until it factors nicely..
k would equal 5.. to give us
x2-5x+4 = (x-1)(x-4)
the (x-1) would cancel .. leaving just x-4.. and the limit would be 1-4 = -3...

Is there an easier way of doing this than just guessing to try to figure out which value of k would make the polynomial factor nicely so that it would cancel with the factor in the denominator? Because, some of the other problems in this section get a little too tough to just be able to spit out the answer...

2. Sep 17, 2006

### Warr

I'm not sure you could say this is easier. But definately more systematic.

For the problem you just solved, you can break the function as such:

$$x^2-kx+4=(x+a)(x+b)=x^2+(a+b)x+ab$$

To cancel the bottom, you know one of the roots must be 1 and therefore must have a (x-1) factor on top. Hence let a=-1. Also from the first equation, you know that ab = 4. Therefore b=-4. From the equation, you know that -k=(a+b)=-5, so k=5.

3. Sep 18, 2006

### Jacobpm64

ok, I understand how to do those now.. But, how would I set up something like this?

lim (e2x - 5) / (ekx + 3)
x-> -infinity

4. Sep 19, 2006

### Warr

Well, try to think about that one logically. What happens to e^2x and e^kx (assuming k is not negative) as x-> -infinity.