Find a when lim f(x)=(x^2+x+a)/(x-1) = 3 x->1

[jgthb]

Hi

I just found an old high school math exercise and was wondering how it should be answered mathematically correct. The exercise is:

Given lim f(x)=(x^2+x+a)/(x-1) = 3 for x->1, determine a. Back in high school I used to say something like this:

Since the denominator goes to 0 for x -> 1 the fraction must be reducible, so x=1 must be a root in the numerator, and thus we can write it as (x-1)(x-b). Hence:

f(x)=(x-1)(x-b)/(x-1)=x-b.

For x->1 we get:

1-b=3 => b=-2, and thus:

(x-1)(x+2)=x^2+x-2,

so a=-2.

The part I think lacks rigour is the argument that the fraction must be reducible. I guess you have to go on about it in some other way. Any suggestions?

/Jonathan

 

[tiny-tim]

Welcome to PF!

Hi Jonathan! Welcome to PF!

(try using the X² tag just above the Reply box )

The part I think lacks rigour is the argument that the fraction must be reducible. I guess you have to go on about it in some other way. Any suggestions?

Looks ok to me … the top must be 0 at x = 1, or the whole thing will be infinite.

But a much quicker way to get a is that if x = 1, the top must be zero, so a = -2.

(More interesting would be to do it for lim_x->1(x² + bx + a)/(x-1) = 3, find a and b )

 

[jgthb]

thanks for the reply.

The problem as I see it is that you cannot look at the limit of the numerator and denominator separately, so to me it seems like these arguments doesn't work. But I think i have figured it out:

If you do the following:

lim((x^2+x+a)/(x-1))=lim((x^2+x+a-2+2)/(x-1))=lim((x-1)(x+2)+(a+2))/(x-1))
=lim((x+2)+(a+2)/(x-1))=lim((x+2))+lim((a+2)/(x-1))=3+lim((a+2)/(x-1)).

Now if a != -2 the limit is indefinite (since it goes towards +infinity for x -> 1+ and -infinity for x -> 1-), and thus the limit is only valid for a=-2.

 

[tiny-tim]

The problem as I see it is that you cannot look at the limit of the numerator and denominator separately, so to me it seems like these arguments doesn't work. But I think i have figured it out:

If you do the following …

Yes, that also works.

But you can look at the limit of the numerator and denominator separately, in that you can say that if the bottom limit is 0, then the top limit must be also, or the result will be ∞.

(in fact, I think that's so obvious, i wouldn't even bother to say it)

 

[jgthb]

when it comes to limits I like to have some theorem to backup intuition. but okay, this time it really is obvious as you say, so I'll just accept it :)