What Are the Values of c and d for This Precalculus Limit Problem?

[Lennie Oswald]

Homework Statement

limit as x approaches 2 of [ (x^2-cx+d)/(x^2-4) ] = 3. What is the value of c and d?

Homework Equations

Limits

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.

 

[Mark44]

Homework Statement

limit as x approaches 2 of [ (x^2-cx+d)/(x^2-4) ] = 3. What is the value of c and d?

Homework Equations

Limits

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.

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.

 

[Molar]

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

 

[Mark44]

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

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.

 

[SammyS]

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 posted the question.

(@Molar isn't OP.)

 

[Lennie Oswald]

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.

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

 

[SteamKing]

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

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.

 

[Lennie Oswald]

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.

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

 

[Lennie Oswald]

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.

Thanks for your help!

 

[SammyS]

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

I get a different sign for one of those.

 

[Ray Vickson]

Homework Statement

limit as x approaches 2 of [ (x^2-cx+d)/(x^2-4) ] = 3. What is the value of c and d?

Homework Equations

Limits

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.

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 $c, d$.