Hi, i have this question which i am unsure whether the answer is 0 or 2.

If x tend to 2-, I know the limit is 2.

How about if x tends to 2+? Please see image:

http://img232.imageshack.us/img232/2...0cma120q10.gif

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- Mar 30th 2010, 07:13 PMtottijohnFinding limit of discontinous function (graph)
Hi, i have this question which i am unsure whether the answer is 0 or 2.

If x tend to 2-, I know the limit is 2.

How about if x tends to 2+? Please see image:

http://img232.imageshack.us/img232/2...0cma120q10.gif - Mar 30th 2010, 08:02 PMProve It
- Mar 30th 2010, 08:20 PMtottijohn
0? But since 0 does not exist, is it 2?

- Mar 30th 2010, 09:38 PMProve It
Yes, the answer is 0.

You are finding out what $\displaystyle f(x)$ APPROACHES as you are making $\displaystyle x$ approach a value. That value of $\displaystyle f(x)$ need not be defined. - Mar 31st 2010, 01:51 AMHallsofIvy
?? 0 certainly does exist(Rofl)! You mean that the value of the function, at x= 2, is not 0. And that is irrelevant to the limit!

If you look carefully at the definition of limit you will see that we require $\displaystyle |f(x)- L|< \epsilon$ only if $\displaystyle 0< |x- a|< \delta$. That "0< |x- a|" means that what happens**at**x= a is irrelevant. It is only what happens**near**the point (and, here, larger) that is important.

It is very, very important to understand that the value of a limit at a point has nothing to do with the value of the function at that point!

For example, f(x)= $\displaystyle x^2- 1$ has limit 3 as x goes to 2. Now define g(x) to be $\displaystyle x^2- 1$ as long as x is NOT equal to 2 and**any**value you want for x= 2. No matter what value you assign**at**x= 2, $\displaystyle \lim_{x\to 2} g(x)= 3$.

Sometimes students get the idea that "limit" is just another way of talking about the value of a function but that's only because "continuous" functions, which are**defined**as functions whose limit is the same as its value, are so easy to use. The function you have here is NOT contininuous at x= 2. In a very specific sense "almost all" functions are not continuous.