Advanced Math/Limit

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Expert: - 6/26/2011


Question
hello
I have learned Limit theory of math well.but I have a basic question and that is why basically determinate form  occurs?

for example if X/X =1 why when we approach to zero 0/0 occurs?
where is the point that we should  evaluate Indeterminate form ?

Answer
Limit and 0/0 occuring

If we have x/x, this is cleary 1 everywhere but at x=0.
However, since it is 1 everywhere else, we can say limit(x->0)[x/x] = 1,
for no how matter how close to 0 we choose the x value, the answer is 1.

For some other examples ...

An example of 0/0 is the limit(x->1)[(x-1)/(x-1)].

This gives 0/0 when x is put in, so there is no value in the function at x=0.
However, since (x=1) is in the numerator and the denominator,
we can see the functioin is just 1, so the limit as x->1 is 1.

If we take limit(x->3)[(x^2 + 4x - 21)/(x-3)],
we also get 0/0 by putting in x=3.

In this case, the top factors into (x-3)(x+7), so we have (x-3)(x+7)/(x-3).
Cancelling the (x-3) in the top and bottom gives x+7, so as we let x goto 3, x+7 goes to 10.
This is a case where we get 0/0, but the limit is not 1.


As far as the intermediate value theorem, it does not specify what point to use,
but just that there is such a point in the interval somewhere.
That is true as long as the function is continous.

If we had the integral from 10 downto 0 of x, the value would be 50.
Note that b-a = 10-0 = 10, so the point to use would be x=5.

If we have the integral from 12 downto 6 of x^2, the integral would be x^3/3.
Now 12^3/3 - 6^3/3 = 576 - 72 = 504.  Since b-a = 12-6 = 6, 504/6 = 84.
The point to use here would be where x^2 = 84, or x = √84 = 2√21.
Since √21 is a little over 4.5, 2 times that is a little over 9.

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