lim [x->2] [(x^2-9)/(x-3)], lim [x->3] [(x^2-9)/(x-3)]

[nima23]

Find lim x-> 2 (x^2 - 9) / (x - 3)

-I got that the limit is at 5 because when you plug in 2 for X, that's what it comes out to.


next problem:

Find lim x-> 3 (x^2 - 9) / (x - 3)

-I got that it DNE because plugging in 3 would make the numerator 0. and when plugging it into the calculator, it says ERROR for x being 3.


did i do these right?

 

[nima23]

Re: Some easy lim problems

or, would it be the opposite... as in where it says ERROR on the calculator, that means the lim does exist and just find that number.
and if it has a value displayed that means it DNE

 

[Pklarreich]

Re: Some easy lim problems

Find lim x-> 2 (x^2 - 9) / (x - 3)

-I got that the limit is at 5 because when you plug in 2 for X, that's what it comes out to.

next problem:

Find lim x-> 3 (x^2 - 9) / (x - 3)

-I got that it DNE because plugging in 3 would make the numerator 0. and when plugging it into the calculator, it says ERROR for x being 3.
did i do these right?

The first is correct; generally if you plug in your x = a, and get a sensible answer, it is probably the correct limit. But ERROR on the calculator just means "Please use pen and paper." The result could be:

1. Does not exist.
2. Can be simplified to something that will ALLOW you to plug it in.

In this case: (x^2 - 9) / (x - 3) = [(x+3)(x-3)]/(x - 3) which simplifies to x + 3. And now you can use your x = 3.

I presume you won't need a calculator to get 6.

There is a lot of theory behind this, but that's why you have a book and a teacher.