suppose a general polygon of some side length x , as we increase the number of sides the polygon would start to look more and more like a circle , and its side length would be limit of x as x approaches 0. however is the limit was equal to 0 the there would not be any circle at all. Hence limit need to be a tiny-bit above 0
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please answer if there was any mistake in this prove
Just like 1+1/2+1/4+1/8... never actually equals 2 when we say limit we mean we can get however close we want to that value. Here too, it never actually equals zero but it can go infinitely close.
Actually I think you are wrong. Imagine a square with are 2 units . Cut it in half to obtain 2 figures of area 1 and 1 then cut one of these figures into half so in total we have 1 +1/2 +1/2 =2 , cut one of the two figures with area 1/2 to obtain 1 +1/2 +1/4+1/4=2 and so on by continuing this process we obtanin 1+1/2+1/4+1/8... =2
//ly the repeating decimal .49999... is actually exactly and precisely equal to .5 - no ifs, ands or buts.
I'm not sure, but it looks like you may be asking how you can obtain the perimeter or area of a circle by using an infinite sequence of regular inscribed n-gons. Yes it is true that as the length of the side --->0 as n--->inf but you have to remember that 1) the distance from the vertices remains constant and will become the radius of the circle and 2) we are summing an infinite number of infinitely small sides around the edge. In the end this infinite sum converges to the perimeter of the circle.
So, if you join 2 adjacent vertices to the center you get an isosceles triangle with sides R,R and delta(s). Remember R is a constant. As n -->inf, delta(s) --> ds ( the arc of the circle) and the contained angle t --->0 since t = 2pi/n.
So ds := Rdt and P= int(ds)
P = int[rdt) from t = 0 to t = 2pi
= rt] 0 to 2pi
= 2piR
Thanks , but I was thinking if the perimeter of shape (polygon ) be equal to circumference or will it APPROACH circumference . More importantly are approach and equal the same as when we evaluate any limit we in end say that limit IS EQUAL TO some value , or there is no limit, but how can something APPROACHING EQUAL TO SOMETHING . When we evaluate limit as x approaches zero , it is very similar to f(0), so why do we need limit st all.
There is a subtle but important difference between these 2 statements:
1) As x --> 0 the expression f(x) = sin(x)/x --> 1 but never takes on the value 1.
2) lim(as x -->0)[sin(x)/x] = 1 and equality is achieved. But it is not equal = f(0) since f(0) is indeterminate. The graph of f(x) = sin(x)/x has a hole at x = 0 even though the limit exists.
i.e. The ---> 1 means we can make sin(x)/x arbitrarily close to 1 by making x sufficiently close to 0 and 1 is the only number that has this property. And "in the limit" equality is achieved. So this answers your question: so why do we need limit st all
As far as the infinite series 1/2 + 1/4 + 1/8 + ... is concerned the "..." means we are adding an infinite number of terms and equality is achieved and the infinite sum actually equals 1.
To say that 1/2 + 1/4 + 1/8 + .... approaches 1 is incorrect.
There is no difference between: sum(from n=1 to inf) [1/2^n] = 1 and 1/2 + 1/4 + 1/8 + ... = 1
The same is true for 9.99999.... = 10. This is exactly the same situation as the above. It is an infinite geometric series which has a definite finite value.
That's the best I can do.
You have to define more acurately the problem. Suppose the polygon is inscribed in a circle of constant radius.
If x is the length of a side and n the number of sides, as n tends to infinity, x tends to 0. If you want to calculate the perimeter just multiplying the two limits, you get an indeterminate form "0 x infinity". The right way is to express x in term of n and the choosen radius, and you get the 2piR formula.
Oh ! I didn't see Ian Fowler's answer before to write mine. Sorry !