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I just found an old highschool 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 highschool 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

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

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