Another Limit Problem

[turophile]

The problem: Let F(k) be the limit as x approaches 4 of the function (x[sup:d16ynqc3]2[/sup:d16ynqc3] - 2x + 2k) / (x[sup:d16ynqc3]2[/sup:d16ynqc3] - 3x + k). Show that F(k) = 2 for all k <> - 4, but that F(- 4) = 6/5.

My work so far: I can show that F(- 4) = 6/5.

Let k = - 4. Then the function whose limit is to be evaluated is (x[sup:d16ynqc3]2[/sup:d16ynqc3] - 2x - 8) / (x[sup:d16ynqc3]2[/sup:d16ynqc3] - 3x - 4).

Now (x[sup:d16ynqc3]2[/sup:d16ynqc3] - 2x - 8) / (x[sup:d16ynqc3]2[/sup:d16ynqc3] - 3x - 4) = [(x - 4)(x + 2)] / [(x - 4)(x + 1)] = (x + 2) / (x + 1) = 1 + 1 / (x + 1).

So F(- 4) = the limit at x approaches 4 of the function 1 + 1 / (x + 1) = 1 + 1 / (4 + 1) = 1 + 1 / 5 = 6/5

But I'm not sure how to start working on the first part of the problem, i.e., showing that F(k) = 2 for all k <> - 4. Any hints or guidance would be appreciated.

 

[galactus]

When x-->4, then F(k)=2. Because if we let x=4, the rational expression reduces to \(\displaystyle \frac{(4)^{2}-2(4)+2k}{(4)^{2}-3(4)+k}=\frac{2k+8}{k+4}=\frac{2(k+4)}{k+4}=2\), regardless of k's value.

But, what value makes \(\displaystyle \frac{8+2k}{4+k}\) undefined?. That's right....\(\displaystyle k = -4\). 8+2k=4+k=0.

There's a kind of 'hole' there when k=-4.

And when k=-4, F(k)=6/5.

\(\displaystyle \frac{x^{2}-2x-8}{x^{2}-3x-4}=\frac{1}{x+1}+1=\frac{x+2}{x+1}\)

\(\displaystyle \frac{1}{4+1}+1=\frac{6}{5}\)