I need HELP! Is there any way to know beforehand whether an undefined limit has a possible answer or not

Don't think my title is clear enough, so let me explain.

I'll have to solve for different kinds of undetermined limits in my exam, but the thing is, I can't differentiate between the limits that have a possible answer and the ones that don't.

Let me illustrate it with an example:

lim (x^2 - 1) / (x-1)
x -> 1

If I substitute 1 for x in this case I get 0/0. However, if I substitute x for very close numbers to 1 (like 1.000001 or 0.999999) I realize this function doesn't have a limit, because those close numbers bigger than 1 go to infinity while those a bit less than 1 go to minus infinity.

Now consider this limit:

lim (2x-2)/(x-1)
x->1

Again, if I substitute x for 1 I get 0/0. Yet this time, when I replace it by very close numbers to 1, I see it has a limit: 2. Actually, if I factor the polynomial ( (2*(x-1))/(x-1) ) I get 2 too.


My question is, how can I know in advance if an undefined limit will actually exist or not? I don't want to waste so much time substituting x with my calculator.

Thanks! 10 points to the BA! :)

1) lim (x^2 - 1) / (x-1)
x -> 1
(x+1)(x-1)/(x-1) = x+1
lim x->1 = 2

2) lim (2x-2)/(x-1)
x->1
2(x-1)/(x-1)
lim x->1 = 2

theres a thing a came up with.

its called undefined nonexistence of 7/4

its (7/4)(doesnotexist/0). *put it in paper to make sense

if the biggest exponent is on the denominator then the limit is 0
if the biggest exponent is on the numerator then the limit does not exist
if the numerator and the denominator have the same biggest exponent the the limit is the constant of the biggest exponent.

examples: (x^3+3)/ x^2+1. in this case limit does not exist

(7x^4 +3)/ 4x^4 +5), the limit is (7/4), i like (7/4) if you have not noticed

No, you cannot tell whether a limit is defined or not without figuring out what the limit is. The method you are trying to use will not work.

The first expression DOES have a limit. It is 2. (Do the division first, and you'll see that clearly.)

The second expression is simply 2*1. Of course it will equal 2 as x approaches 1. It will equal 2 under any circumstances.

The fact that your method erroneously distinguished between two expressiins which each have a limit is a sign it's not working.

I think you'd better review some solved problems. There are lots available if you google them.