When is a limit considered to exist?

[Jimmy25]

Homework Statement

Find a value of k such that the limit exists

lim (x²-kx+4)/(x-1)
x->1

The Attempt at a Solution

In the solution they set the top equal to zero finding that k is equal to 5. Why do you assume the numerator must equal zero if the denominator equals zero? Is this the only case in which the limit exists? If so why?

 

[ideasrule]

If the top equals anything else, the fraction (x^2-kx+4)/(x-1) would blow up near x=1 because the denominator approaches 0. Setting the top to 0 ensures that x-1 is a factor of the numerator. The x-1 on the top would then cancel with the denominator, preventing the fraction from becoming infinity near x=1.

 

[Jimmy25]

So in this case, they are assuming that an infinite limit does not exist?

 

[Mark44]

They are implicitly saying that a finite limit exists.

 

[HallsofIvy]

So in this case, they are assuming that an infinite limit does not exist?

I would not use the word "assuming" here- it is a fact.

Saying that a limit "is infinity" or "is negative infinity" is just saying that the limit does not exist for a specific reason.