Find any singularities in the folloqing fucntion

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andrey21
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Find any singularities in the following function, say whether they are removable or non-removable. Indicate the limit of f(x) as x approaches the singularity.

(x^(2) + x + 1) /( x-1)


Not to sure where to start, as the numerator does not factorise easily.
 
on Phys.org
Sure the numerator factorizes easily. It's a quadratic equation. Just calculate the discriminant [tex]D=b^2-4ac[/tex] and the roots [tex]\frac{-b\pm\sqrt{D}}{2a}[/tex]...

That said, you don't really need to factorize the numerator. A singularity is just a point in the domain where the function is not defined. What could be the problem with the function such that it is not defined??
 
If a function is not defined will there be no singularities??
 
Ok so the place where the function is not defined is x^2 + x + 1: So the singularity is non-removable?
 
andrey21 said:
Ok so the place where the function is not defined is x^2 + x + 1: So the singularity is non-removable?

No, that doesn't even make sense :frown: You need to find out in what specific point the function is not defined. For example, is the function defined in 5? That is, can you calculate the function value of 5? (in this case: yes!). Are there any specific values that you cannot calculate the value of??
 
Ok so when we have f(1) this gives:

1^(2) + 1 + 1 / (1-1)

3/0

Which isn't defined!
 
Ok so would this singularity be non-removable?
 
A removable singularity is a point at which the function is undefined.
 
Im sorry that is the only definition I have.
 
In simple terms, a removable singularity is a singularity that you can "remove". For example:

[tex]f(x) = x^2/x[/tex] has a removable singularity at x = 0, since:

[tex]x^2/x = x[/tex] for all x not equal to 0, and we can define f(0) = 0. How can we apply this same idea to your problem?

(Now if you want the technical definition, a removable singularity is a point at which the function is undefined, but it is possible to define the function at that point in such a way that the function is regular in a neighbourhood of that point. It doesn't look like we need a lot of detail in this problem though)
 
Ok so f(1) = 3/0 so would this suggest it is a non-removable singularity??
 
Correct. It doesn't look like this singularity is removable since the function is unbounded near x = 1.
 
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