Expert: Daniel Mazur - 7/10/2009

Question
Hi Daniel,

Wanted to discuss the idea of limits.
I understand the idea but still sort of feel a little uncomfortable.

Let's take an example:

let f(x) = (x^2 - 4)/(x - 2)

If we wish to find the limit at the point x=2,

we take points close to 2 on either side of 2.
For example I took a point 2.01 and put it in the above expression and got the answer as 4.01.
Also, this expression can be written down as just (x+2) after cancelling out the common factor (x-2).

So we can equivalently write the function as x+2 for points lying in the immediate neighbourhood of x=2.

But when we put x=2 in the expression, we get 0/0.
My first question is, what does 0/0 mean? Does it mean that we don't know what it means??(!)
Also, if we put x=2 in x+2, we get the answer as 4.
Why is that??
And why are the two expressions, i.e. (x^2 - 4)/(x - 2) and x+2 not equivalent??
And if they are, (I can't figure out why they aren't)
why do we get different answers on x=2 for the two expressions??

And if they aren't equivalent, then how come we get the same and correct answer, by using any of the two, for points in the neighbourhood of 2??


Cheers!!
Shikhin


Answer
Shikhin,

the difference between (x^2 - 4)/(x - 2) and (x + 2) is ONLY that the former expression does not have a defined value for x=2, while the latter has one. This is precisely because 0/0 is an non-evaluable expression - it could be ANYTHING. We are only allowed to formally divide by (x-2) for x NOT equal to 2. Even if in this case we would be safe to do so (for the purposes of physics anyway), there are many other examples that are not as transparent as this and therefore we must follow the rigorous mathematical route - use the limits.

You need to study limits properly, it doesn't do the job to just dip in and get mystified... :-) The purpose of limits is varied, but one of them is that if we come across an expression like your f(x) without a defined value at some x, we can smoothly extend the expression to cover all points including x=2. Apart from the fact that your f(x) can be reduced to (x + 2), there is a standard procedure do calculate a value of a point limit, it's called the L'Hospital Rule:

lim{x->2} (x^2 - 4)/(x - 2) = lim{x->2} [d(x^2 - 4)/dx] / [d(x - 2)/dx]

Simply the limit of a ratio is equal to the limit of the ratio of DERIVATIVES (der. of numerator divided by der. of denominator) by parameter, in which we consider the limit (x here). This rule can be proven, naturally, but don't ask the proof from me here - I would have to look it up. I am sure it is in all first year's college maths books. In any case, when you apply the rule to your f(x), you get

lim{x->2} (x^2 - 4)/(x - 2) = lim{x->2} 2x/1 = lim{x->2} 2*x

And of course the right-hand limit is nothing other than 2*2 = 4, which equals to your substitution to the reduced formula (x+2) = 4.

Cheers,
Daniel