# Help on precalculus hw question

1. Aug 29, 2016

### Lennie Oswald

1. The problem statement, all variables and given/known data
limit as x approaches 2 of [ (x^2-cx+d)/(x^2-4) ] = 3. What is the value of c and d?

2. Relevant equations
Limits

3. The attempt at a solution
I tried cross multiplying to get (x^2 - cx + d) = 3x^2 - 12, and I was thinking of moving combining like terms and maybe factoring but I can't ever seem to get the right answer.

2. Aug 29, 2016

### Staff: Mentor

The denominator factors into (x - 2)(x + 2). If you assume that the numerator also has a factor of either x + 2 or x - 2, what would the other factor have to be to result in a limit of 3? An educated guess or two will be helpful.

3. Aug 29, 2016

### Molar

@Mark44
Just for curiosity, can we apply L' hospital's rule here, assuming the numerator to be 0 when x→2 ?

4. Aug 29, 2016

### Staff: Mentor

I didn't think of doing that, but it seems to work. That technique wouldn't be available in a precalculus setting, which is what I assumed by where you the OP posted the question.

Last edited: Aug 29, 2016
5. Aug 29, 2016

### SammyS

Staff Emeritus
(@Molar isn't OP.)

6. Aug 29, 2016

### Lennie Oswald

I tried that but I want to know how to solve it algebraically.

7. Aug 29, 2016

### SteamKing

Staff Emeritus
I don't think that's possible.

In the limit, the denominator $x^2 - 4$ goes to 0 as x → 2.

Cross multiplying means that you are multiplying the RHS by zero as well, leaving $x^2 - cx + d = 0$, for which there can be an infinite number o' solutions.

Evaluating limit expressions sometimes takes subtlety, where applying algebra will fail or mislead you.

8. Aug 29, 2016

### Lennie Oswald

Thanks! I ended up getting c = 8 and d = -20

9. Aug 29, 2016

### Lennie Oswald

10. Aug 29, 2016

### SammyS

Staff Emeritus
I get a different sign for one of those.

11. Aug 30, 2016

### Ray Vickson

It is easiest to substitute $x = 2 + h$ in both the numerator and denominator. Your ratio must make sense as $h \to 0$, and the limit must = 3. That gives you two conditions involving the two parameters $c, d$.