Can anyone give me a good definition a of a limit?

Question

on a related note how is the series 1/2 + 1/4 +1/8..... converging on 1?

Great Answers! So basically, do linear functions not have limits and higher functions do?

Answer 1

The definition of a limit is slightly different depending on whether you are looking at the limit of a function at some point\, or the limit of a function at infinity, or the limit of a sequence at infinity. I will do the last.

We say that the limit of the sequence a_n as n goes to infinity is the number L if whenever e>0, there is a natrual number N so that whenever n>=N, we have |a_n -L|
In other words, the limit is L if for every 'tolerance' e, we can go far enough out in the sequence (N) so that every term of the sequence farther out is within the tolerance of the purported limit.

As an example. Let a_n=1/2 +1/4 +...1/2^n.
It is easy to show by induction that a_n =1 - 1/2^n.

Given a tolerance e>0, choose N so that 1/e <2^N. Then, if n>=N,
we have |a_n -1|=1/2^n < e.

This shows that the limit L is 1. It has to be proved that limits are unique when they exist, but that is not too difficult. Any good advanced calculus course will cover this stuff.

Answer 2

The definition of a limit is slightly different depending on whether you are looking at the limit of a function at some point\, or the limit of a function at infinity, or the limit of a sequence at infinity. I will do the last.

We say that the limit of the sequence a_n as n goes to infinity is the number L if whenever e>0, there is a natrual number N so that whenever n>=N, we have |a_n -L|
In other words, the limit is L if for every 'tolerance' e, we can go far enough out in the sequence (N) so that every term of the sequence farther out is within the tolerance of the purported limit.

As an example. Let a_n=1/2 +1/4 +...1/2^n.
It is easy to show by induction that a_n =1 - 1/2^n.

Given a tolerance e>0, choose N so that 1/e <2^N. Then, if n>=N,
we have |a_n -1|=1/2^n < e.

This shows that the limit L is 1. It has to be proved that limits are unique when they exist, but that is not too difficult. Any good advanced calculus course will cover this stuff.

OR

First lets look at your specific problem and then at limits in general

The sum of an infinite series is defined to be the limit of the partial sums, if the limit exists. In your example, the sequence of partial sums is 1/2, 3/4, 7/8, 15/16, ... in general the nth partial sum will be (2^n-1)/2^n = 1 - 1/2^n.

The definition of a limit of a sequence says that if z is the limit of the sequence a_1, a_2, ... then given any number, usually designated by epsilon, there is an number N such that for n>N, the distance between a_n and z is less than epsilon.

Pu ting that in English, given any number (distance), no matter how small, if you go out in the sequence far enough, all the terms from there on out will be less than that distance from the limit value.

In your example, given an epsilon, you can pick N such that 1/2^N < epsilon. So the distance between (1-1/2^n) and 1 will be less than epsilon when n>N.

With functions, you can look at the limit of f(x) as x goes to a finite value or as x goes to infinity. The limit as x goes to infinity definition is pretty much the same as the one for sequences: If you are give any small number, eventually, if you pick x large enough f(x) stays close (within epsilon) of the limit.

If you look at limits, at a finite value, the definition goes like this:

The limit as x->a of f(x) = z if and only if, given epsilon >0 there exists delta > 0 such that |f(x) - z| < epsilon or all x such that |x-a|
Again in English, f(x) may not be defined for x = a, but if you
you can force f(x) to b3e close to z by picking a small enough interval around a.

Answer 3

First lets look at your specific problem and then at limits in general

The sum of an infinite series is defined to be the limit of the partial sums, if the limit exists. In your example, the sequence of partial sums is 1/2, 3/4, 7/8, 15/16, ... in general the nth partial sum will be (2^n-1)/2^n = 1 - 1/2^n.

The definition of a limit of a sequence says that if z is the limit of the sequence a_1, a_2, ... then given any number, usually designated by epsilon, there is an number N such that for n>N, the distance between a_n and z is less than epsilon.

Pu ting that in English, given any number (distance), no matter how small, if you go out in the sequence far enough, all the terms from there on out will be less than that distance from the limit value.

In your example, given an epsilon, you can pick N such that 1/2^N < epsilon. So the distance between (1-1/2^n) and 1 will be less than epsilon when n>N.

With functions, you can look at the limit of f(x) as x goes to a finite value or as x goes to infinity. The limit as x goes to infinity definition is pretty much the same as the one for sequences: If you are give any small number, eventually, if you pick x large enough f(x) stays close (within epsilon) of the limit.

If you look at limits, at a finite value, the definition goes like this:

The limit as x->a of f(x) = z if and only if, given epsilon >0 there exists delta > 0 such that |f(x) - z| < epsilon or all x such that |x-a|
Again in English, f(x) may not be defined for x = a, but if you
you can force f(x) to b3e close to z by picking a small enough interval around a.

Answer 4

Limit is something like the "squeezing theorem"

If you have a number, a, and a function f(x). As x gets closer to a from the positive or negative side, then f(x) gets closer to a particular number, L. L is then said to be the limit.

as n becomes large without bound, sum[ (1/2)ⁿ] =1. basically, you take a half-step toward 1 with each new term.

-edit-

You later asked if linear functions have limits. Sure they do. Take y=3x+9. lim(x->4) of y is 21. Just plug in the number and go.

Now take y=(3x²+9x) / x. This function is undefined for x=0 due to the x in the denominator. Otherwise, it's exactly like y=3x+9. So, lim (x->0) of y=3x²+9x is still 9.

Answer 5

A limit is a value that a graph is approaching. It might not be the value of the function at a point due to a hole or a piecewise function.

1/2=.5
1/2+1/4=.75
1/2+1/4+1/8=.875
1/2+1/4+1/8+1/16=.9375

see how its getting closer and closer to one but will never go beyond?

Answer 6

Pick a number between 1 and 10.

Then on your convergence question... in increments of half of the preceeding number until it hits '1'. Which will be never.

Answer 7

A limit is a number which a sequence/series or a function APPROACHES but does not necessarily have to be defined at that particular number.

Answer 8

Greatest or smallest amount permissible....Out of Webster

Answer 9

Advanced math has never been one of my strong points unfortunately

Answer 10

Twelve drinks.